Answer: A
Explanation: The mass of a thing never changes but weight is the act of gravity on mass. This rules out B and C since mass can’t change. Leaving A as the only possible answer.
A hockey puck moving at 0.44 m/s collides elastically with another puck that was at rest. The pucks have equal mass. The first puck is deflected 39° to the right and moves off at 0.34 m/s. Find the speed and direction of the second puck after the collision.
The speed and direction of the second puck after the collision are 0.44 m/s to the right. Let's consider the first puck that was moving at 0.44 m/s before the collision and after the collision moves at 0.34 m/s and at an angle of 39° to the right. We can calculate the velocity vectors of the two pucks before and after the collision, as well as the momentum vectors before and after the collision.
The momentum and velocity vectors can be calculated as follows: Puck 1 initial velocity: v₁ = 0.44 m/s to the right. Puck 1 initial momentum: p₁ = m₁v₁Puck 1 final velocity: v₁' = 0.34 m/s at 39° to the rightPuck 1 final momentum: p₁' = m₁v₁'Puck 2 initial velocity: v₂ = 0 m/s. Puck 2 initial momentum: p₂ = m₂v₂Puck 2 final velocity: v₂'Puck 2 final momentum: p₂' Using the law of conservation of momentum, we can say that:p₁ + p₂ = p₁' + p₂'Therefore, since both pucks have equal mass, m₁ = m₂=p₁ = p₁' + p₂' The x-component of the momentum is conserved since there are no external forces acting in the horizontal direction. p₁x = p₁'x + p₂'xp₁x = m₁v₁ cosθ₁p₁'x = m₁v₁' cosθ₁'p₂'x = m₂v₂' cosθ₂'θ₁ = 0° (initial direction is to the right)θ₁' = 39° (final direction is to the right and up)θ₂' = θ₁' - 90° = -51° (final direction is to the left and up). Therefore,p₁x = p₁'x + p₂'xm₁v₁ = m₁v₁' cosθ₁' + m₂v₂' cosθ₂'m₁v₁ = m₁v₁' cos39° + m₂v₂' cos(-51°)The mass of the pucks is equal so we can simplify this equation to:v₁ = v₁' cos39° + v₂' cos(-51°)Substituting the given values,0.44 m/s = 0.34 m/s cos39° + v₂' cos(-51°)Solving for v₂',v₂' = (0.44 m/s - 0.34 m/s cos39°)/cos(-51°) = 0.44 m/s to the right (rounded to two significant figures)
Hence, the speed and direction of the second puck after the collision are 0.44 m/s to the right.
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During dry conditions, a hiker climbs from 5300 ∘
to 6000 ∘
. At 5300 ′
, the temperature is 60F. What is the most likely femperature at 6000 ? Provide your answer in F (no unit, just the number).
The temperature at 6000 is likely to be 53°F. The reason is that as one climbs up the mountain, the temperature decreases by approximately 3.5°F every 1000 feet of elevation gain.
Here, the elevation gain is 700 feet, so the temperature is expected to drop by around 24.5°F (700/1000 × 3.5). Therefore, if the temperature is 60°F at 5300 feet, it is expected to be 60°F - 24.5°F = 35.5°F lower at 6000 feet.
A hiker climbing from 5300 ft to 6000 ft during dry conditions can expect a change in temperature. The temperature difference arises due to the difference in elevation between the two points. As the hiker gains elevation, the temperature generally decreases. To determine the temperature at the top of the climb, one can use the estimated rate of temperature drop per unit elevation gain.
On average, the temperature drops by about 3.5°F per 1000 feet of elevation gain. The elevation gain in this problem is 700 feet (6000-5300), so the temperature change can be estimated to be -24.5°F (700/1000 x -3.5°F).
Since the temperature at 5300 feet is given to be 60°F, we can subtract the change in temperature from the starting temperature to find the most likely temperature at 6000 feet. The resulting temperature is 60°F - 24.5°F = 35.5°F. Therefore, the most likely temperature at 6000 feet is 35.5°F.
The temperature at 6000 is expected to be 53°F, as the elevation difference between the two points is 700 feet and the temperature usually drops by around 3.5°F every 1000 feet of elevation gain. As a result, we can conclude that if the temperature is 60°F at 5300 feet, it is expected to be 60°F - 24.5°F = 35.5°F lower at 6000 feet.
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A 33.5-g glass thermometer reads 21.6°C before it is placed in 139 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 42.8°C. Ignore the mass of fluid inside the glass thermometer. The value of specific heat for water is 4186 J/kg.Cº, and for glass is 840 J/kg.Cº. What was the original temperature of the water? Express your answer using three significant figures.
The original temperature of the water was approximately 29.7°C. The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature.
To solve this problem, we can use the principle of energy conservation. The energy gained by the water will be equal to the energy lost by the thermometer.
The energy gained by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
where:
m_water is the mass of the water,
c_water is the specific heat capacity of water, and
ΔT_water is the change in temperature of the water.
The energy lost by the thermometer can be calculated using the formula:
Q_thermometer = m_thermometer * c_thermometer * ΔT_thermometer
where:
m_thermometer is the mass of the thermometer,
c_thermometer is the specific heat capacity of glass, and
ΔT_thermometer is the change in temperature of the thermometer.
Since the thermometer and the water come to equilibrium, the energy gained by the water is equal to the energy lost by the thermometer:
Q_water = Q_thermometer
m_water * c_water * ΔT_water = m_thermometer * c_thermometer * ΔT_thermometer
Rearranging the equation, we can solve for the initial temperature of the water (T_water_initial):
T_water_initial = (m_thermometer * c_thermometer * ΔT_thermometer) / (m_water * c_water) + T_water_final
Given:
m_water = 139 g (converted to kg)
c_water = 4186 J/kg.Cº
ΔT_water = 42.8°C - 21.6°C = 21.2°C
m_thermometer = 33.5 g (converted to kg)
c_thermometer = 840 J/kg.Cº
ΔT_thermometer = 42.8°C - T_water_initial
Substituting these values into the equation, we can solve for T_water_initial:
T_water_initial = (0.0335 kg * 840 J/kg.Cº * (42.8°C - T_water_initial)) / (0.139 kg * 4186 J/kg.Cº) + 21.6°C
Simplifying the equation, we get:
T_water_initial = (0.0335 * 840 * 42.8) / (0.139 * 4186) + 21.6
Calculating the right-hand side of the equation, we find:
T_water_initial ≈ 29.7°C
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An m=0.4 kg ball is dropped straight down from the top of a building and strikes the ground after t=1.7 s. Friction is negligible. Find the speed just before the ball strikes the ground.
An m=0.3 kg ball is thrown horizontally at vi=3.2 m/s from the top of a building and falls for t=3.2 s. Friction is negligible, consider only after the throw. Find the final velocity.
You push your biology textbook, m=1.1 kg, across your desk with an initial vi=1.7 m/s until it comes to rest in t=1.3 s. Find the average resistance force.
The speed just before the ball strikes the ground is 16.66 m/s, the final velocity of the thrown ball is 35.36 m/s, and the average resistance force on the textbook is -1.43 N.
For the first scenario, the speed just before the ball strikes the ground can be found using the equation v = gt, where g is the acceleration due to gravity. Since the ball is dropped, the initial velocity is 0. By substituting the values, we find v = (9.8 m/s²)(1.7 s) = 16.66 m/s.
For the second scenario, the final velocity can be determined using the equation v = vi + gt, where vi is the initial horizontal velocity and g is the acceleration due to gravity. Since the ball is thrown horizontally, the vertical initial velocity is 0. By substituting the values, we have v = 3.2 m/s + (9.8 m/s²)(3.2 s) = 35.36 m/s.
For the third scenario, the average resistance force can be calculated using the equation F = (mΔv) / Δt, where m is the mass of the textbook, Δv is the change in velocity, and Δt is the change in time. The change in velocity is given by Δv = vf - vi, where vf is the final velocity and vi is the initial velocity. Substituting the values, we find Δv = 0 - 1.7 m/s = -1.7 m/s. Then, the average resistance force is F = (1.1 kg)(-1.7 m/s) / 1.3 s = -1.43 N.
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The half-life of 131
I is 8.04 days. (a) Convert the half-life to units of seconds. 5 (b) What is the decay constant (in s −1
) for this isotope? s −1
(c) Suppose a sample of 131
I has an activity of 0.460 uCi. What is this activity expressed in the 51 unit of becquerels (Bq)? Bq (d) How many 131
I nuclei are needed in the sample in part (c) to have the activity of 0.460μci ? 131
1 nuclei (e) Now suppose that a new sample of 131
thas an activity of 6.70mCl at a given time. How many half-lives will the sample go through in next 40.2 days? (Enter your answer for the number of half-lives to at least one decimal place., half-lives What is the activity of this sample (in mCl) at the end of 40.2 days?
(a) Half-life is the time taken for half the number of nuclei in a sample of an isotope to decay. The half-life of 131I is 8.04 days. To convert half-life into units of seconds:Half-life = 8.04 days = 8.04 × 24 × 60 × 60 seconds = 693,504 seconds ≈ 150 × 60 × 60 seconds.
(b) The decay constant (λ) of 131I is calculated as follows:λ = 0.693 ÷ t1/2λ = 0.693 ÷ 693504λ = 1 × 10−6 s−1(c) Activity is the rate of decay of a sample. The activity of the sample of 131I is 0.460 μCi. 1 μCi = 37,000 Bq, then 0.460 μCi = 0.460 × 37,000 Bq = 17,020 Bq(d) To calculate the number of 131I nuclei needed in the sample in part (c) to have the activity of 0.460 μCi, use the following equation:Activity = decay constant × number of nucleiN0 = Activity ÷ (decay constant)N0 = 17020 ÷ (1 × 10−6)N0 = 17.02 × 106(e) To calculate the number of half-lives the sample of 131I will go through in the next 40.2 days, use the following equation:t1/2 = (ln2) ÷ λλ = (ln2) ÷ t1/2λ = 0.693 ÷ 8.04λ = 8.61 × 10−2 day−1After 40.2 days, the number of half-lives is:τ = (40.2 days) ÷ (8.04 days/half-life)τ = 5 half-lives.The activity of this sample (in mCi) at the end of 40.2 days can be calculated using the following equation:N = N0 × (1/2)τN = 17.02 × 106 × (1/2)5N = 1.064 × 106The activity of 131I is expressed as:Activity = decay constant × number of nuclei × 37The decay constant (λ) of 131I is calculated as follows:λ = 0.693 ÷ t1/2λ = 0.693 ÷ 693504λ = 1 × 10−6 s−1Activity = 1 × 10−6 × 1.064 × 106 × 37 = 39.2 mCi at the end of 40.2 days.
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A 14 V battery delivers 104 mA of current when connected to a 74 Ω resistor. Determine the internal resistance of the battery. Answer in units of Ω.
The internal resistance of the battery is 60.5 Ω (approx).
Voltage of battery (V) = 14 V
Current passing through it (I) = 104 mA = 0.104 A
Resistance of the resistor (R) = 74 Ω
To find the internal resistance of the battery, use the formula;
Voltage of battery (V) = Current passing through it (I) × (Resistance of the resistor (R) + Internal resistance of the battery (r))
Putting the above values in the formula we get:
14 V = 0.104 A × (74 Ω + r)
14 V = 7.696 Ω + 0.104 r
0.104 r = 14 V - 7.696 Ω
0.104 r = 6.304 Ω
r = 6.304 / 0.104 Ω
r = 60.5 Ω (approx)
Therefore, the internal resistance of the battery is 60.5 Ω (approx).
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If you drive with a constant velocity of 24 m/s East for 4s, what would your acceleration be during this time? 6 m/s^2 0 m/s2 20 m/s^2 96 m/s^2
If a vehicle maintains a constant velocity of 24 m/s East for 4 seconds, the acceleration during this time would be [tex]0 m/s^2[/tex].
Acceleration is the rate at which an object's velocity changes. In this scenario, the vehicle is moving with a constant velocity of 24 m/s East. Since velocity remains constant, there is no change in velocity, and therefore the acceleration is [tex]0 m/s^2[/tex].
Acceleration is only present when there is a change in velocity, either in terms of speed or direction. In this case, since the vehicle maintains a steady speed and travels in a straight line without any change in direction, there is no acceleration occurring. Acceleration would only be present if the vehicle were to speed up, slow down, or change its direction. Therefore, the correct answer is [tex]0 m/s^2[/tex].
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While on safari, you see a cheetah 10 m away from you. The cheetah starts running at t= 0. As it runs in a straight line away from you, its displacement can be described as x(t) = 10 m+ (5.0 m/s2)ť. (a) Draw a graph of the cheetah's displacement vs. time. х t (b) What is the average velocity of the cheetah during the first 4 seconds of its run? (c) What is the average velocity of the cheetah from t = 4.9 s to t= 5.1 s? (d) What is the instantaneous velocity of the cheetah at any time t? In other words, what is v(t)? (e) How does your answer for (C) compare to the instantaneous velocity at t= 5.0 s?
(a) The cheetah's displacement vs. time, the equation is x(t) = 10 m + [tex](5.0 m/s^2[/tex])t. (b) The average velocity during the first 4 seconds can be calculated by finding the change in displacement (Δx) divided by the change in time (Δt). (c) The average velocity from t = 4.9 s to t = 5.1 s can be calculated in the same way. Δx = x(5.1 s) - x(4.9 s) and Δt = 5.1 s - 4.9 s.
(d) The instantaneous velocity, v(t), at any time t can be found by taking the derivative of the displacement function x(t) with respect to time. In this case, v(t) = dx(t)/dt = d/dt (10 m + ([tex]5.0 m/s^2[/tex])t). (e) To compare the average velocity at t = 5.0 s to the instantaneous velocity, we can calculate the instantaneous velocity at t = 5.0 s .
(a) The displacement vs. time graph of the cheetah will be a straight line with a positive slope of [tex]5.0 m/s^2[/tex] The initial displacement at t = 0 s is 10 m, and the displacement increases linearly with time due to the constant acceleration of [tex]5.0 m/s^2[/tex].
(b) To find the average velocity during the first 4 seconds, we need to calculate the change in displacement (Δx) during that time interval and divide it by the change in time (Δt). This gives us the average rate of change of displacement, which is the average velocity. By substituting the values into the formula, we can find the average velocity during the first 4 seconds.
(c) Similarly, to find the average velocity from t = 4.9 s to t = 5.1 s, we calculate the change in displacement (Δx) during that time interval and divide it by the change in time (Δt). This gives us the average velocity during that specific time interval.
(d) The instantaneous velocity at any time t can be found by taking the derivative of the displacement function with respect to time. In this case, we differentiate x(t) = 10 m + ([tex]5.0 m/s^2[/tex])t with respect to t, giving us the instantaneous velocity function v(t) = [tex]5.0 m/s^2[/tex].
(e) To compare the average velocity at t = 5.0 s to the instantaneous velocity, we substitute t = 5.0 s into the instantaneous velocity function obtained in part (d). By comparing this value to the average velocity calculated in part (c), we can determine how they differ or coincide.
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An object in SHM oscillates with a period of 4.0 s and an amplitude of 13 cm. Part A How long does the object take to move from x = 0.0 cm to x = 5.5 cm. Express your answer with the appropriate units
We need to express our answer with appropriate units, which is seconds (s).The answer is 0.449 s.
Given,Period of oscillation T = 4.0 sAmplitude A = 13 cmThe equation of motion of an object in SHM is given as:x = A sin (ωt)where, A = Amplitudeω = Angular frequency (ω = 2π/T)Therefore, the equation becomes:x = A sin (2π/T * t)For finding time period of oscillation, we need to find angular frequency first:ω = 2π/T = 2π/4.0 = π/2 rad/sx = A sin (ωt)x = 13 sin (π/2 * t)At maximum displacement, i.e. x = 5.5 cm13 sin (π/2 * t) = 5.5sin (π/2 * t) = 5.5/13
Let's solve the above equation to get the time of oscillationt = (1/π)sin-1(5.5/13) = 0.449 sTherefore, the object takes 0.449 seconds to move from x = 0.0 cm to x = 5.5 cm.However, we need to express our answer with appropriate units, which is seconds (s).Thus, the answer is 0.449 s.
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A 120 g block slides down an incline plane of inclination angle 25 ∘
. A . 25 N friction force impedes the sliding motion of the block. Find the magnitude of the acceleration undergone by the block. Select one: a. 2.06 m/s 2
b. 2.89 m/s 2
C. 1.17 m/s 2
d. 1.89 m/s 2
e. 3.54 m/s 2
The magnitude of the acceleration undergone by the 120 g block sliding down an incline plane with an inclination angle of 25° and a friction force of 0.25 N is approximately 2.89 m/s².
To find the magnitude of the acceleration, we need to consider the forces acting on the block. The gravitational force (mg) acts vertically downward, and the friction force (f) opposes the motion.
The component of the gravitational force along the incline plane is given by mg sin(θ), where θ is the inclination angle. The net force (F_net) acting on the block can be expressed as:
[tex]\[ F_{\text{net}} = m \cdot a \][/tex]
[tex]\[ F_{\text{net}} = mg \sin(\theta) - f \][/tex]
Substituting the given values, where m = 0.120 kg, g ≈ 9.8 m/s², θ = 25°, and f = 0.25 N:
[tex]\[ 0.120 \cdot a = (0.120 \cdot 9.8 \cdot \sin(25\°)) - 0.25 \][/tex]
Simplifying the equation:
[tex]\[ a = \frac{(0.120 \cdot 9.8 \cdot \sin(25\°)) - 0.25}{0.120} \][/tex]
[tex]\[ a \approx 2.89 \, \text{m/s²} \][/tex]
Therefore, the magnitude of the acceleration undergone by the block is approximately 2.89 m/s². Thus, the correct option is (b) 2.89 m/s².
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You just realized that your analog wristwatch is always 25 seconds behind the real-time. Calculate the angular speed of your Second hand, in milli-ads/s. A 26 kg skip attached to a steel rope on a crane is used to hoist bricks from the ground to the top of a construction site. The steel rope is wound onto a lifting drum with a diameter of 700 mm and rotational frequency of 56 revolutions per minute. The lifting drum is situated on the top floor which is 195 m high. How many seconds will it take to lift bricks, three quarters up the height of the building?
The angular speed of the second hand is 104.67 milli-radians/s.
The drum will take approximately 16.92 seconds to lift bricks three-quarters up the height of the building.
Analog watch is 25 seconds behind the real-time.
Rotational frequency of lifting drum is 56 revolutions per minute.
Diameter of the lifting drum is 700 mm.
The lifting drum is situated on the top floor which is 195 m high.
The mass of the skip is 26 kg.
Conversion factor: 1 minute = 60 s.
Angular speed of the second hand:We know that the time period of the watch is 60 seconds. The time period (T) is the time taken by an object to complete one revolution.
So, Angular speed (ω) = 2π / T = 2π / 60 rad/s = π / 30 rad/s
In milli-radians per second, angular speed = (π / 30) × 10³ milli-radians/s = 104.67 milli-radians/s (approx.)
The length of the steel rope = 195 m. The mass of the skip is 26 kg.
So, Total work done = mgh
where m = mass of the skip = 26 kg
g = acceleration due to gravity = 9.8 m/s²
h = height to which bricks are lifted = (3 / 4) × 195 m = 146.25 m
Total work done = 26 × 9.8 × 146.25 J
Total work done = 37,617 J
The diameter of the lifting drum = 700 mm.
So, Radius of the drum, r = 700 / 2 = 350 mm = 0.35 m
Rotational frequency (n) = 56 rev/min = 56 / 60 rev/s = 0.9333 rev/s
Circumference of drum, C = 2πr = 2 × π × 0.35 = 2.1991 m
The distance traveled by the rope in one revolution of the drum = circumference of drum = 2.1991 m
The distance traveled by the rope in one revolution of the drum = 2.1991 m
Energy required to lift the skip one time = Total work done / efficiency
where efficiency = 90% = 0.9
Work done by the rope in one revolution = energy required / efficiency
Work done by the rope in one revolution = 37,617 J / 0.9 = 41,797 J
The work done by the rope in one revolution of the drum is equal to the work done in lifting the skip one time.
Distance covered by the rope in one revolution of the drum = 2.1991 m
Work done by the rope in one revolution of the drum = 41,797 J
So, the force applied to lift the skip = Work done / Distance = 41,797 / 2.1991 = 19,000 N
The time taken to lift the skip three-quarters of the height of the building can be calculated as follows:Height to which the skip is lifted, h = 146.25 m
Let's say the skip is lifted a distance x at time t.
Since the force is constant, the distance is proportional to time.
x / t = F / m(g - a)
where g = acceleration due to gravity = 9.8 m/s²
a = acceleration of the skip = (F / m)
Distance left to lift the skip = h - x
The initial velocity of the skip = 0 m/s
The final velocity of the skip = vf
Time taken, t = (vf - vi) / a
The final velocity can be calculated using the kinematic equation:
v² - u² = 2as
where u = initial velocity = 0 m/s
v² = 2as
Therefore, v = √(2as)
The acceleration of the skip = (F / m) - g.
a = (F / m) - g
Let's substitute the known values in the equations:
x / t = F / m(g - a)
x / t = F / m(g - (F / m) + g)
x / t = F² / ma
Let's substitute the value of acceleration in the above equation:
x / t = F² / m((F / m) - g)
x / t = F² / (mg - F²)
The height to which the skip is lifted, h = 146.25 m.
The skip is lifted three-quarters of this height. Therefore,
x = 3h / 4 = 109.6875 m
Let's substitute this value in the above equation:
109.6875 / t = F² / (mg - F²)
Let's substitute the known values in the above equation:
109.6875 / t = (19,000)² / (26 × 9.8 - (19,000)²)
109.6875 / t = 361,000,000 / 3,044,000
109.6875t = 30.85
t ≈ 0.282 minutes = 16.92 s
Therefore, it will take approximately 16.92 seconds to lift bricks three-quarters up the height of the building.
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Consider the two point charges shown in the figure below. Let q1 =(-5)×10–6 C and q2=1×10–6 C.
A) Find the x-component of the total electric field due to q1 and q2 at the point P.
B) Find the y-component of the total electric field due to q1 and q2 at the point P.
C) Find the magnitude of the net electric force due to q1 and q2 on an electron placed at point P.
D) Find the angle that the net electric force due to q1 and q2 on an electron placed at point P makes with the positive x-axis.
A) The x-component of the total electric field due to q1 and q2 at the point P is 15.28 × 10⁶ N/C.B)The y-component of the total electric field due to q1 and q2 at the point P is 10.18 × 10⁶ N/C.C)The magnitude of the net electric force due to q1 and q2 on an electron placed at point P is 2.59 × 10⁻¹³ N.D)The angle that the net electric force due to q1 and q2 on an electron placed at point P makes with the positive x-axis is 33.18°.
A) The x-component of the total electric field due to q1 and q2 at the point P is given by Ex = E1x + E2x, where E1x and E2x are the x-components of the electric fields due to charges q1 and q2 respectively.So, Ex = E1x + E2x = k × (q1/r1²)cosθ1 + k × (q2/r2²)cosθ2where k is Coulomb's constant, r1 and r2 are the distances between the point P and charges q1 and q2 respectively, θ1 and θ2 are the angles made by the lines joining the charges to the point P with the positive x-axis.
The negative sign before q1 indicates that the direction of the force on an electron due to a positive charge is opposite to the direction of the electric field at the point.
Pictorial representation of the situation is given below:As per the right-angled triangle formed in the above diagram:θ1 = tan⁻¹(4/3) = 53.13°θ2 = tan⁻¹(3/4) = 36.87°So, Ex = k × (q1/r1²)cosθ1 + k × (q2/r2²)cosθ2= (9 × 10⁹ × 5 × 10⁻⁶/5²) cos(53.13°) + (9 × 10⁹ × 1 × 10⁻⁶/4²) cos(36.87°)= 15.28 × 10⁶ N/C.
Thus, the x-component of the total electric field due to q1 and q2 at the point P is 15.28 × 10⁶ N/C.
B) The y-component of the total electric field due to q1 and q2 at the point P is given by Ey = E1y + E2y, where E1y and E2y are the y-components of the electric fields due to charges q1 and q2 respectively.So, Ey = E1y + E2y = k × (q1/r1²)sinθ1 + k × (q2/r2²)sinθ2where k is Coulomb's constant, r1 and r2 are the distances between the point P and charges q1 and q2 respectively, θ1 and θ2 are the angles made by the lines joining the charges to the point P with the positive x-axis.
The negative sign before q1 indicates that the direction of the force on an electron due to a positive charge is opposite to the direction of the electric field at the point.
Pictorial representation of the situation is given below:As per the right-angled triangle formed in the above diagram:θ1 = tan⁻¹(4/3) = 53.13°θ2 = tan⁻¹(3/4) = 36.87°So, Ey = k × (q1/r1²)sinθ1 + k × (q2/r2²)sinθ2= (9 × 10⁹ × 5 × 10⁻⁶/5²) sin(53.13°) + (9 × 10⁹ × 1 × 10⁻⁶/4²) sin(36.87°)= 10.18 × 10⁶ N/C.
Thus, the y-component of the total electric field due to q1 and q2 at the point P is 10.18 × 10⁶ N/C.
C) The magnitude of the net electric force due to q1 and q2 on an electron placed at point P is given by Fnet = qE, where q is the charge of the electron and E is the net electric field at point P.Fnet = qE = -1.6 × 10⁻¹⁹ × √(Ex² + Ey²)Fnet = -2.59 × 10⁻¹³ N.
Thus, the magnitude of the net electric force due to q1 and q2 on an electron placed at point P is 2.59 × 10⁻¹³ N.
D) The angle that the net electric force due to q1 and q2 on an electron placed at point P makes with the positive x-axis is given by θ = tan⁻¹(Ey/Ex)θ = tan⁻¹(10.18 × 10⁶ / 15.28 × 10⁶)θ = 33.18°.
Thus, the angle that the net electric force due to q1 and q2 on an electron placed at point P makes with the positive x-axis is 33.18°.
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A spring-block system sits on a horizontal, frictionless surface. The spring has a spring constant k = 295 N/m. The mass of the block is 6.7 kg. The spring is stretched out and released at t=0.00 s. The block undergoes simple harmonic motion. if the magnitude of the block's acceleration at t= 2.9 s is 13.4 cm/s², determine the total energy (mJ) of the spring-block system?
Answer: the total energy (mJ) of the spring-block system is 1.00 mJ.
mass of the block m = 6.7 kg
Spring constant k = 295 N/m
Initial position of the block = 0 (because the spring is stretched).
The block undergoes simple harmonic motion. The magnitude of the block's acceleration at t = 2.9 s is a = 13.4 cm/s² = 0.134 m/s².
The total energy (mJ) of the spring-block system can be found using the formula for total mechanical energy, E which is E = 1/2 kA²
E = 1/2 mv² + 1/2 kx²
whereA is the amplitude. v is the velocity of the block at a particular instant of time x is the displacement of the block from its equilibrium position. The total energy of the spring-block system can be found as follows; We know that the block undergoes simple harmonic motion and the magnitude of the block's acceleration at
t = 2.9 s is a = 13.4 cm/s² = 0.134 m/s².
The displacement of the block from its equilibrium position at t = 2.9 s can be found using the formula for the displacement of the block, x which is x = Acosωt where A is the amplitudeω is the angular frequency t is the time. The angular frequency can be found using the formula,ω = √k/m. Substituting k = 295 N/m and m = 6.7 kg,ω = √(295/6.7) rad/s = 6.09 rad/s. Substituting ω = 6.09 rad/s, t = 2.9 s and A = x/ cos ωt13.4 cm/s² = Aω²cos ωt.
Therefore, A = 0.0751 m. The total energy of the spring-block system can be found using the formula for total mechanical energy, E which isE = 1/2 kA²E = 1/2 x 295 x (0.0751)²E = 1.00 mJ.
Therefore, the total energy (mJ) of the spring-block system is 1.00 mJ.
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A car is initially traveling along a highway at vo–30 m/s. A truck, which is S-10 meter away in front of the car, is also traveling along the highway at the same spoed vo^30 mv's in the same direction at the side lane. Atr-o the car begins to accelerate at a constant acceleration in order to pass the truck. It takes the car ty -2 seconds to pass the truck. Please calculate the acceleration of the car. a- (Please provide your numerical answer without unit! Please write your numerical answer with all digits and do not use scientific notation. If you are not sure about the number of significant figures, you can keep the number of figures as many as possible - You will not be punished for doing this. No unit in your answer.)
The acceleration of the car is 35 m/s².
Given:
Speed of car initially, vo = vo - 30 m/s
Speed of truck, vo = vo
Speed of car after passing truck = vo + 30 m/s
Distance between car and truck, S = 10 m
Time taken by car to pass the truck, t = 2 s
To calculate:
Acceleration of the car, a
We can use the formula:
S = ut + 1/2 at^2
Here, initial velocity, u = vo - 30 m/s,
final velocity, v = vo + 30 m/s, and
distance, S = 10 m.
We need to calculate the acceleration,
a.
By substituting the given values in the above formula, we get:
S = (vo - 30) × 2 + 1/2 a(2)^2
Simplifying this we get:
10 = 2vo - 60 + 2aOn
simplifying this we get:
2a = 70 - 2voa = 35 - vo
We know that, vo = vo - 30
So, a = 65 - vo
Substituting vo = 30 m/s in the above equation,
we get:
a = 35 m/s²
Therefore, the acceleration of the car is 35 m/s².
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Suppose 2000 J of heat are added to 3.4 mol of argon gas at a constant pressure of 140 kPa. Part A Find the change in internal energy. Part B Find the change in temperature for this gas. Express your answer using two significant figures
Part C Calculate the change in volume of the gas.
The resulting change in temperature of the argon gas is approximately 34.62 Kelvin.
To determine the change in temperature of the argon gas, we can use the formula:
ΔQ = nCpΔT
where:
ΔQ is the heat added to the gas (in joules),
n is the number of moles of the gas,
Cp is the molar specific heat capacity of the gas at constant pressure (in joules per mole per kelvin),
ΔT is the change in temperature (in kelvin).
In this case, we have:
ΔQ = 2000 J
n = 3.4 mol
Cp (specific heat capacity of argon at constant pressure) = 20.8 J/(mol·K) (approximately)
We need to rearrange the formula to solve for ΔT:
ΔT = ΔQ / (nCp)
Substituting the given values into the equation, we have:
ΔT = 2000 J / (3.4 mol * 20.8 J/(mol·K))
Calculating the result:
ΔT ≈ 34.62 K
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--The complete Question is, Suppose 2000 J of heat are added to 3.4 mol of argon gas at a constant pressure of 140 kPa. What will be the resulting change in temperature of the gas? Assume the argon gas behaves ideally.--
A toaster is rated at 660 W when connected to a 220 V source. What current does the toaster carry? A. 2.0 A B. 2.5 A C. 3.0 A D. 3.5 A
The given toaster is rated at 660 W when it is connected to a 220 V source. We can find the current that the toaster as follows,
P = VI or I=P/V, where P is the power, V is the voltage, I is the current
So, I=660/220
I=3A
Therefore, the current that the toaster carries C. 3.0 A.
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A 120 V, 50 Hz, 0.50 Hp, Two-Pole, Resistance Split-Phase Induction Motor Has The Following Main Winding Impedances: Z1 = (1.72 + J2.65) Ω Z2 = (2.36 + J2.65) Ω XM = 90 Ω PF&W = 35 W For A Slip Of 0.05 P.U, Determine: 1.The Magnitude Stator Current In Amps 2.For A Slip Of 0.05 P.U, Determine: The Magnitude Stator Current In Amps 3.The
A 120 V, 50 Hz, 0.50 hp, two-pole, resistance split-phase induction motor has the following main winding impedances:
Z1 = (1.72 + j2.65) Ω
Z2 = (2.36 + j2.65) Ω
XM = 90 Ω
PF&W = 35 W
For a slip of 0.05 p.u, determine:
1.The magnitude stator current in amps
2.For a slip of 0.05 p.u, determine: The magnitude stator current in amps
3.The input power in watts
4.Air-gap power in watts
The correct answer is 1) Magnitude of I1 = |I1| = 1.22 A 2) 1.22 A. 3) 4.85 Wb. and 4) 354 W.
1. The magnitude stator current in amps:
Given data:
Voltage, V = 120V
Frequency, f = 50 Hz
Output power, Pout = 0.50 hp
Slip, S = 0.05
Let the current flowing through stator winding is I1
Now the rotor input power Pinput is given by,
Pinput = Pout / efficiency = Pout / (Pout + losses)
For a two-pole induction motor,
Pinput = (Pout + Pf & W + Pg)
Where Pf & W is friction and windage loss and Pg is the air-gap power.
Now, Pout = 0.50 hp × 746 W/hp = 373 W
Pg = Pout (1 - S) = 373(1 - 0.05) = 354 W
Pf & W = 35 W (Given)
Pinput = (373 + 35 + 354) = 762 W
So, the stator input power Pin is,
Pin = Pinput / ω = Pinput / (2πf)
where ω is the angular velocity of the rotating magnetic field.ω = 2πf / P = 2π × 50 / 2 = 157.08 rad/sec
Pin = 762 / 157.08 = 4.85 Wb
Let's calculate the stator current. For that, we need to calculate the total impedance Z_total as
Z_total = Z1 + Z2 + jXM
= (1.72 + j2.65) + (2.36 + j2.65) + j90
= 4.08 + j95.3 Ω
The current through stator winding is given as,
I1 = V / Z_total
I1 = 120 / (4.08 + j95.3)
I1 = 1.22 ∠ -87.8° A
Magnitude of I1 = |I1| = 1.22 A (Ans)
2. For a slip of 0.05 p.u, determine: The magnitude stator current in amps:
We have already calculated the magnitude of the stator current in part 1, which is equal to 1.22 A.
3. The input power in watts:
The input power to the motor is calculated in part 1 which is equal to 4.85 Wb.
4. Air-gap power in watts:
The air-gap power is calculated in part 1 which is equal to 354 W.
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A whetstone of radius 4.0 m is initially rotating with an angular velocity of 89 rad/s. The angular velocity is then increased at 10 rad/s for the next 12 seconds. Assume that the angular acceleration is constant. What is the magnitude of the angular acceleration of the stone (rad/s2)? Give your answer to one decimal place
The magnitude of the angular acceleration of the stone is 0.8 rad/s² (rounded to one decimal place).
Radius of the whetstone (r) = 4.0 m
Initial angular velocity (ω₀) = 89 rad/s
Change in angular velocity (Δω) = 10 rad/s
Time interval (t) = 12 s
The final angular velocity (ω) can be calculated as:
ω = ω₀ + Δω
Substituting the given values:
ω = 89 + 10 = 99 rad/s
To find the angular acceleration (α), we use the formula:
α = Δω / t
Substituting the values:
α = 10 / 12 ≈ 0.8 rad/s²
Therefore, the magnitude of the angular acceleration of the stone is 0.8 rad/s² (rounded to one decimal place).
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A gun is fired vertically into a block of wood (mass ml) at rest directly above it. If the bullet has a mass of m2 and a speed of v, how high will the block rise into the air after the bullet becomes embedded in it?
Answer: the height to which the block will rise into the air after the bullet becomes embedded in it is given by
H = (m₂v)² / 2(m₁ + ml)g.
When a gun is fired vertically into a block of wood at rest directly above it, the velocity of the block can be calculated by applying the law of conservation of momentum. Here, the bullet of mass m₂ is fired into the block of wood of mass ml. According to the law of conservation of momentum, the initial momentum of the bullet and the final momentum of the bullet and the block combined must be equal, and it can be expressed as:m₂v = (m₁ + ml)VWhere V is the velocity of the bullet and the block combined.
From the equation, we have: V = m₂v / (m₁ + ml)As the bullet and the block rise to a maximum height H, their total energy is equal to their initial kinetic energy, given as: 1/2 (m₁ + m₂) V² = (m₁ + m₂)gh. Where g is the acceleration due to gravity. Solving for H, we get: H = V² / 2g
Substituting the value of V in the above equation, we have: H = (m₂v)² / 2(m₁ + ml)g.
Therefore, the height to which the block will rise into the air after the bullet becomes embedded in it is given by H = (m₂v)² / 2(m₁ + ml)g.
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A dentist's drill starts from rest. After 2.70 s of constant angular acceleration, it turns at a rate of 2.51×10 4
rev/min. (a) Find the drill's angular acceleration. rad/s 2
(along the axis of rotation) (b) Determine the angle through which the drill rotates during this period. rad
(a) Angular acceleration is 972.9 [tex]rad/s^2[/tex] (b) angle through which the drill rotates during this period is 3520.8 rad.
The rate at which the angular velocity of an item changes over time is determined by its angular acceleration. It measures the rate of change in rotational speed or direction of an object. The difference between the change in angular velocity and the change in time is known as angular acceleration.
It is measured in radians per square second (rad/s2) units. An increase in angular velocity is indicated by positive angular acceleration, whereas a decrease is indicated by negative angular acceleration. It is affected by things like the torque that is given to an object, that object's moment of inertia, and any outside forces that are acting on it. Understanding rotational motion and the behaviour of rotating objects requires an understanding of angular acceleration, a fundamental term in rotational dynamics.
(a) The formula for the angular acceleration is given by the following:α = ωf - ωi/t
The given values are,ωi = 0 (The drill starts from rest)ωf = 2.51×104 rev/min = (2.51×104 rev/min)*([tex]2\pi[/tex] rad/1 rev)*(1 min/60 s) = 2628.9 rad/st = 2.70 sα = ?
Therefore,α = (2628.9 rad/s - 0 rad/s)/(2.70 s)α = 972.9 rad/[tex]s^2[/tex]
Therefore, the angular acceleration of the drill is 972.9 rad/[tex]s^2[/tex].
(b) The formula for the angular displacement is given by the following:θ = ωi*t + (1/2)α[tex]t^2[/tex]
The given values are,ωi = 0 (The drill starts from rest)t = 2.70 sα = 972.9 rad/[tex]s^2[/tex]
Therefore,θ = 0*(2.70 s) + [tex](1/2)*(972.9 rad/s²)*(2.70 s)²θ[/tex] = 3520.8 rad
Therefore, the angle through which the drill rotates during this period is 3520.8 rad.
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Retake question A 4.5 Kg package of kiwi flavored bubble gum is being delivered to the ground floor of an office building. The box sits on the floor of an elevator which accelerates downward with an acceleration of magnitude a=-3.0 m/s².The delivery person is also resting one foot on the package exerting a downward force on the package of magnitude 5.0 N. What is the normal force on the package exerted by the floor of the elevator. 63 N 36 N 126 N 31 N
Substituting the given values, we getN = F - ma= 5.0 N - (4.5 kg)(-3.0 m/s²)= 5.0 N + 13.5 N= 18.5 N.Therefore, the normal force exerted on the package by the floor of the elevator is 18.5 N.
Given:Mass of package, m= 4.5 kg Downward acceleration, a = -3.0 m/s²Downward force exerted by delivery person, F = 5.0 N Let N be the normal force exerted on the package by the floor of the elevator.Thus, the equation of motion for the package along the downward direction isF - N = ma.Substituting the given values, we getN = F - ma= 5.0 N - (4.5 kg)(-3.0 m/s²)= 5.0 N + 13.5 N= 18.5 NTherefore, the normal force exerted on the package by the floor of the elevator is 18.5 N.
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Two pistons of a hydraulic lift have radii of 2.67 cm and 20.0 cm. A mass of 2.00×10^3 kg is placed on the larger piston. Calculate the minimum downward force needed to be exerted on the smaller piston to hold the larger piston level with the smaller piston.
------------- N
The minimum downward force needed to be exerted on the smaller piston to hold the larger piston level with the smaller piston is 348.8 N.
A hydraulic lift works on Pascal’s principle which states that pressure applied to an enclosed fluid is transmitted equally in all directions. The pressure applied to the fluid is equal to the force applied per unit area. A hydraulic lift system consists of two pistons of different sizes connected by a pipe filled with fluid. The force applied on one piston gets transmitted to the other piston with a force that is multiplied by the ratio of the area of the two pistons.
The area of the smaller piston is given as follows:A = πr²where r = 2.67 cm = 0.0267 mTherefore, A = π(0.0267)² = 0.002232 m²The area of the larger piston is given as follows:A = πr²where r = 20.0 cm = 0.20 mTherefore, A = π(0.20)² = 0.1257 m²Since the force exerted on the larger piston is due to the weight of the mass placed on it, we can calculate the force as follows:F = mgwhere m = 2.00×10³ kg, and g = 9.81 m/s²Therefore, F = (2.00×10³)(9.81) = 19.62 kN = 1.962×10⁴ N.To calculate the minimum downward force needed to hold the larger piston level with the smaller piston, we can use the ratio of the area of the two pistons. Let F₁ be the force needed to be exerted on the smaller piston.
Therefore, the force exerted on the larger piston is given as:F₂ = F₁ × (A₂ / A₁)where A₁ is the area of the smaller piston, and A₂ is the area of the larger piston.Since the two pistons are at the same level, the force exerted on the larger piston is equal and opposite to the force exerted on the smaller piston. Therefore, we can write:F₁ = F₂ / (A₂ / A₁)F₁ = (1.962×10⁴) / (0.1257 / 0.002232)F₁ = 348.8 NTherefore, the minimum downward force needed to be exerted on the smaller piston to hold the larger piston level with the smaller piston is 348.8 N.
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Three resistors are connected in parallel. If their respective resistances are R1 = 23.0 Ω, R2 = 8.5 Ω and R3 = 31.0 Ω, then their equivalent resistance will be: a. 5.17 Ω
b. 62.5 Ω
c. 0.193 Ω
d. 96.97 Ω
The equivalent resistance of the three resistors connected in parallel is 5.17 Ω.
The equivalent resistance of the three resistors that are connected in parallel is calculated as follows:
The formula for calculating the equivalent resistance for resistors in parallel is given as:
1/Rp = 1/R1 + 1/R2 + 1/R3 +...+ 1/Rn
where Rp is the equivalent resistance, and R1, R2, R3 and so on are the resistances in ohms.
The values of resistances are given as:
R1 = 23.0 Ω
R2 = 8.5 Ω
R3 = 31.0 Ω
Substitute the given values of resistances into the equation:
1/Rp = 1/23.0 + 1/8.5 + 1/31.0
1/Rp = 0.043 + 0.118 + 0.032
1/Rp = 0.193
To find the equivalent resistance, we take the reciprocal of both sides of the equation:
Rp = 1/0.193
Rp = 5.18 Ω
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A 0.900 kg hammer is moving horizontally at 4.50 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact. X S (b) What was the average force exerted on the nail? N
The duration of the impact can be calculated by considering the work-energy theorem, while the average force exerted on the nail is calculated by dividing the change in momentum by the duration of the impact.
The hammer comes to a stop after driving the nail 1.00 cm into the board. This implies that it decelerated uniformly. We can use the equation of motion v^2 = u^2 - 2as to find the deceleration, where v is the final velocity (0 m/s), u is the initial velocity (4.50 m/s), a is the acceleration, and s is the distance (1.00 cm = 0.01 m). Solving for a, we get a = (v^2 - u^2) / -2s = -1012.5 m/s^2.
(a) The duration of the impact can be calculated using the equation t = (v - u) / a, resulting in t = -0.00444 seconds (4.44 ms).
(b) The average force exerted on the nail is equal to the change in momentum of the hammer divided by the time taken. The initial momentum is the mass of the hammer times its initial velocity (0.900 kg * 4.50 m/s = 4.05 kg.m/s). The final momentum is zero (as the hammer comes to rest). The change in momentum (Δp) is therefore -4.05 kg.m/s. The average force (F) can then be calculated by dividing this change in momentum by the time of impact, F = Δp / t, which results in -912 N.
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How can astronomers make observations that penetrate the cosmic dust which limits our view of the Galaxy through the disk? By using ultraviolet telescopes. By using radio telescopes. By observing globular clusters. Question 21 If an O-type star and our galaxy were at the same distance, which would have a greater magnitude? The O-type star The Galaxy Insufficient information to say Question 22 Which of the below is a possible evolutionary outcome for the Sun (given in the correct chronological order). planetary nebula, red giant, white dwarf Red giant, planetary nebula, white dwarf Red giant, planetary nebula, neutron star Red giant, neutron star with simultaneous supernova explosion Red giant, black hole with simultaneous supernova explosion
Astronomers make observations that penetrate the cosmic dust which limits our view of the Galaxy through the disk by using radio telescopes as it is one of the best ways to peer into the universe.
Radio telescopes can easily penetrate the cosmic dust in the galaxy, making it possible for astronomers to see through the dust and view the galaxy in ways that are impossible to see with visible light.Radio telescopes allow astronomers to see through gas and dust that are present between the stars and galaxies.
They are used to study the universe's radio waves. Radio telescopes pick up the radio waves emitted by stars and galaxies, and these waves can be used to create images of the universe.
Astronomers use a variety of techniques to peer through the cosmic dust that limits our view of the galaxy through the disk. Radio telescopes are one of the best ways to peer into the universe. Radio telescopes can easily penetrate the cosmic dust in the galaxy, making it possible for astronomers to see through the dust and view the galaxy in ways that are impossible to see with visible light.Radio telescopes are used to study the universe's radio waves. Radio waves are emitted by many objects in the universe, including stars and galaxies. Radio telescopes pick up these waves and use them to create images of the universe.
The images produced by radio telescopes are often more detailed than those produced by visible light telescopes because radio waves are less affected by the dust and gas present in the universe.Globular clusters are also used to study the universe. Globular clusters are large groups of stars that are located outside the Milky Way galaxy. These clusters are some of the oldest objects in the universe and provide astronomers with valuable information about the early universe.
Observing globular clusters allows astronomers to study the chemical makeup of the universe and the conditions that existed in the early universe.If an O-type star and our galaxy were at the same distance, the O-type star would have a greater magnitude. This is because O-type stars are very bright and emit a lot of light.
Magnitude is a measure of the brightness of a star, with brighter stars having a lower magnitude. O-type stars are among the brightest stars in the universe, so they have a lower magnitude than the Milky Way galaxy.The possible evolutionary outcome for the Sun, given in the correct chronological order, is planetary nebula, white dwarf. As the Sun gets older, it will eventually expand into a red giant.
After that, it will shrink down into a planetary nebula and eventually a white dwarf. Planetary nebulae are formed when a red giant sheds its outer layers of gas and dust. The remaining core of the star becomes a white dwarf, which is a very dense object that emits a small amount of light and heat.
Astronomers use different techniques to study the universe and peer through cosmic dust. Radio telescopes, globular clusters, and other methods are some of the ways astronomers use to study the universe. Magnitude is a measure of the brightness of a star, with brighter stars having a lower magnitude.
O-type stars are the brightest stars in the universe. The possible evolutionary outcome for the Sun is planetary nebula and white dwarf, as it expands into a red giant before it shrinks down into a planetary nebula and eventually a white dwarf.
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Charles Cansado launched a 100 g dart upwards from a height of 150 cm using a toy gun. The stiffness of the gun's spring is 1 000 N/m which was compressed 10 cm. Determine the impact velocity of the dart the instant it reaches its target at a height of 450 cm if the heat loss was 0.588 J. Determine the percentage efficiency of the shot.
The impact velocity of the dart when it reaches its target at a height of 450 cm is 5.20 m/s. The percentage efficiency of the shot is 95.2%.
In order to determine the impact velocity of the dart, we can use the principle of conservation of mechanical energy. The initial potential energy of the dart is given by mgh, where m is the mass of the dart (0.1 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (1.5 m). The final potential energy of the dart is mgh, where h is the final height (4.5 m). The initial kinetic energy of the dart is zero, as it was launched from rest. Therefore, the final kinetic energy of the dart is equal to the difference between the initial potential energy and the heat loss (0.588 J). Using these values, we can calculate the final velocity of the dart using the equation KE = 0.5mv^2, where KE is the kinetic energy, m is the mass of the dart, and v is the velocity.
The percentage efficiency of the shot can be determined by calculating the ratio of the actual energy output (final kinetic energy) to the theoretical maximum energy output (initial potential energy). The efficiency is then multiplied by 100 to express it as a percentage. In this case, the efficiency is 95.2%. This means that 95.2% of the energy stored in the spring was transferred to the dart as kinetic energy, while the remaining 4.8% was lost as heat.
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Consider the figure below. (a) Find the total Coulomb force (in N) on a charge of 9.00nC located at x=4.50 cm in part (b) of the figure, given that q=6.50μC. (Indicate the direction with the sign of your answer.) N (b) Find the x-position (in cm, and between x=0 cm and x=14 cm ) at which the electric field is zero in part (b) of the figure. x=cm
(a) The total Coulomb force (in N) on a charge is F = 0.090 NThe direction of the force is repulsive as the two charges are both positive.(b) The x-position where the electric field is zero is 8.22 cm.
(a) The formula for Coulomb's law is:F = (1/4πε) * (q1 * q2 / r²)where ε = permittivity of free space = 8.85 × 10−12 N−1 m−2 C²F = force in Nq1 = 9.00 nCq2 = 6.50 μC = 6.50 × 10−6CThe distance between the charges can be found from the diagram to be:r = 8.0 cm + 4.5 cm = 12.5 cm = 0.125 m.
Therefore, plugging in the values in Coulomb's law equation:F = (1/4πε) * (q1 * q2 / r²)F = (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (9.00 × 10−9C) * (6.50 × 10−6C) / (0.125m)²F = 0.090 NThe direction of the force is repulsive as the two charges are both positive.
(b) To find the x-position at which the electric field is zero, we can use the concept of electric potential.The electric potential at any point due to a point charge is given by:V = (1/4πε) * (q / r)where r = distance between the charge and the point where potential is to be found.
For charges distributed along an axis (as in this case), we can add up the potentials due to all the charges.To find the point where the electric field is zero, we can imagine a positive test charge being placed at different positions along the axis and find at which point the test charge does not experience any force.
The potential at a point on the x-axis at distance x from the first charge q1 is:V1 = (1/4πε) * (q1 / x)V2 = (1/4πε) * (q2 / (14cm - x))At the point where the electric field is zero, V1 + V2 = 0Substituting the given values:V1 + V2 = (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (9.00 × 10−9C) / x + (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (6.50 × 10−6C) / (14cm - x)= 0.
Solving this equation gives the value of x as 8.22 cm (rounded off to two decimal places).Therefore, the x-position where the electric field is zero is 8.22 cm.
Part (a)The force between two point charges is given by Coulomb's Law. The formula for Coulomb's law is:F = (1/4πε) * (q1 * q2 / r²)where F = force in Nε = permittivity of free space = 8.85 × 10−12 N−1 m−2 C²q1 = 9.00 nCq2 = 6.50 μC = 6.50 × 10−6Cr = 8.0 cm + 4.5 cm = 12.5 cm = 0.125 mTherefore, plugging in the values in Coulomb's law equation:F = (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (9.00 × 10−9C) * (6.50 × 10−6C) / (0.125m)²F = 0.090 NThe direction of the force is repulsive as the two charges are both positive.
Part (b)The potential at a point on the x-axis at distance x from the first charge q1 is:V1 = (1/4πε) * (q1 / x)V2 = (1/4πε) * (q2 / (14cm - x))At the point where the electric field is zero, V1 + V2 = 0Substituting the given values:V1 + V2 = (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (9.00 × 10−9C) / x + (1/4π(8.85 × 10−12 N−1 m−2 C²)) * (6.50 × 10−6C) / (14cm - x)= 0.
Solving this equation gives the value of x as 8.22 cm (rounded off to two decimal places).Therefore, the x-position where the electric field is zero is 8.22 cm.
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A 10-A current flows through the wire shown. What is the magnitude of the magnetic field due to a 0.3\( \mathrm{mm} \) segment of wire as measured at: a. point \( A \) ? Magnetic field at A is T. (Use
The magnitude of the magnetic field due to a 0.3 mm segment of wire as measured at point A is 3.2×10−4 T.
Given that:Current flowing through the wire is 10 ALength of the wire is 0.3 mmTo calculate the magnetic field at point A, we can use the Biot-Savart law which states that the magnetic field at a point due to a current-carrying wire is directly proportional to the current flowing through the wire and the length of the wire segment as measured from the point. The formula for magnetic field is given byB=μ0I4πRWhereμ0 = magnetic constant = 4π×10−7 T⋅m/IA = distance of the point from the wireI = current flowing through the wireR = radius of the loop.
Through the given figure, we can see that distance between point A and the wire is 0.6 cm (as given in figure). Therefore, we need to convert it into meters as μ0 is in terms of T⋅m/IMagnetic field at point A due to the wire can be calculated asB = μ0I/2πrB = (4π×10−7)×10/2×3.14×0.006B = 3.2×10−4 TTherefore, the magnitude of the magnetic field due to a 0.3 mm segment of wire as measured at point A is 3.2×10−4 T.
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A diverging lens has a focal distance of -5cm. a) Using the lens equation, find the image and size of an object that is 2cm tall and it is placed 10cm from the lens. [5 pts] b) For the object in 2a) above, what are the characteristics of the image, real or virtual, larger, smaller or of the same size, straight up or inverted?
A diverging lens has a focal distance of -5cm. The focal length of the lens = -5 cm .characteristics of the image will be: Virtual image . Therefore, the image is 3cm tall.
The given diverging lens has a focal distance of -5 cm, and an object of 2cm tall is placed 10cm from the lens.
We need to find the image and the size of the object by using the lens equation.
Lens equation is given as: 1/v - 1/u = 1/f Where ,f is the focal length of the lens, v is the image distance, u is the object distance
Here, the focal length of the lens = -5 cm
Object distance = u = -10 cm (Negative sign indicates the object is in front of the lens)Height of the object = h = 2 cm
Let's calculate the image distance(v) by substituting the values in the lens equation.1/v - 1/-10 = 1/-5Simplifying the equation, we get, v = -15 cm
Since the image distance(v) is negative, the image is virtual, and the characteristics of the image will be: Virtual image
Larger than the object (since the object is placed beyond the focal point)Erect image (since the object is placed between the lens and the focal point)
Therefore, the image is 3cm tall.
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3. All about Ceiling temperature a) What is "ceiling temperature" of a polymerization reaction? (5 pts) b) Explain the relationship between monomer concentration versus its ceiling temperature? (10 pt
Ceiling temperature is defined as the temperature at which the rate of the forward reaction equals that of the backward reaction in a polymerization reaction. This refers to the maximum temperature beyond which polymerization does not proceed, indicating that the polymerization rate is zero at this temperature.
Polymerization reactions are concentration-dependent, which means that they can be significantly influenced by the concentration of monomers. The ceiling temperature, therefore, is directly proportional to the monomer concentration. When the monomer concentration increases, the ceiling temperature also increases. For example, when the concentration of monomers is low, the ceiling temperature of a polymerization reaction is also low, which limits the reaction rate.However, as the concentration of monomers increases, the ceiling temperature of the reaction also increases, allowing for higher reaction rates. As a result, the ceiling temperature plays a critical role in determining the concentration of monomers required for a successful polymerization reaction.The relationship between monomer concentration and ceiling temperature is critical because it helps to establish the ideal conditions for the polymerization reaction. If the concentration of monomers is too low, the ceiling temperature will also be too low, and polymerization will not proceed. Conversely, if the concentration of monomers is too high, the ceiling temperature will also be too high, leading to uncontrolled polymerization reactions. Therefore, understanding the relationship between monomer concentration and ceiling temperature is crucial for optimizing polymerization reactions.
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