Answer:
No
Explanation:
Acceleration is not a "Force"
The only things added on a force diagrams are forces that are acting on the object. Force causes acceleration to change but it is not a force itself and is not included on a force diagram.
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Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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Another nervous papa bear paces at 1.0 m/s north for 3.0 s, then at 1.6 m/s south for 5.0 s, and then
at 1.4 m/s north for 4.0 s. The average speed of this bear is:
a) 1.3 m/s
b) 1.4 m/s
c) 4.0 m/s
d) 8.0 m/s
Answer:
Explanation:
Average speed is total distance traveled over the time needed to do so.
d = vt
v = (1.0(3.0) + 1.6(5.0) + 1.4(4.0) / (3.0 + 5.0 + 4.0)
v = 16.6 / 12
v = 1.383333333...
v = 1.4 m/s
help meee please please please
a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?
Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
At a construction site, a 68.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 85.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 46.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.
Following are the calculation to the friction force:
Please see the diagram below for an illustration of the forces acting on objects.Therefore, m denotes the mass of a gravel bag, and M denotes the overall mass of the box.Whenever the system is stationary, the friction force on the box equals the tension in the string.The substring tension force is provided as follows:
[tex]\to T-m_{1}g=0 \\\\\to T = m_{1}g \\\\[/tex]
[tex]= (68\ kg) (9.81 \ \frac{m}{s^2} ) (\frac{ 1\ N}{ 1\ kg \cdot \frac{m}{s^2}}) \\\\ = (68 ) (9.81 ) ( 1\ N) \\\\= 667.08\ N \\\\[/tex]
Therefore, the friction force on the box is "667 N".
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builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
Four year old Sam has been hitting other children at preschool. He may have learned this behavior from his parents who regularly spank him. What approach to personality best explains Sam's behavior?
A. psychodynamic
B. behavioral
C. humanistic
D. biological
A. psychodynamic is the personality best explains Sam's behavior
What is psychodynamic ?
The psychology of mental or emotional forces or processes developing especially in early childhood and their effects on behavior and mental states is called psychodynamic .
Since , any behavior is not genetical or it never get inherited hence it is not biological . This has been grown with time and this behavior have been grown in Sam by observing his parents doing the same hence correct option should be A. psychodynamic
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Calculate the altitude above the surface of Earth, in meters, at which the acceleration due to gravity is g
Answer:
By definition the acceleration due to gravity at the surface is g:
The altitude above the surface is zero for an acceleration of g.
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
Explanation:
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Two point particles, with masses m and 2 m, are separated by some distance Lin an uniform gravitational field. The center of gravity of this two particle system is:________
А. located L/2 distance away from either mass and on the line joining the two particles.
B. located 1/3 distance away from 2 m and on the line joining the two particles.
C. located L/4 distance away from 2 m and on the line joining the two particles.
D. located L/3 distance away from mand on the line joining the two particles.
E. located L/4 distance away from m and on the line joining the two particles.
F. located where the mass 2 m is located.
The center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
The given parameters:
Mass of the first particle, = mMass of the second particle, = 2mDistance between the two particles, = LThe center of gravity of the two particles when first particle is fixed is calculated as;
[tex]C_G = \frac{m(0 ) \ +2m(L) }{m+ 2m} \\\\C_G = \frac{2mL}{3m} \\\\C_G = \frac{2L}{3}[/tex]
The center of gravity of the two particles when second particle is fixed is calculated as;
[tex]C_G = \frac{m(L) \ + \ 2m(0)}{m + 2m} \\\\C_G = \frac{mL}{3m} \\\\C_G = \frac{L}{3}[/tex]
Thus, the center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
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If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+
Answer:
Explanation:
I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was
Conservation of momentum
14000(8.1) + m(0) = (14000 + m)(3.6)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
The mass of the second car is 17500 kg.
What is law of conservation of momentum?The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Newton's third law of motion has a direct impact on the idea of momentum conservation.
mass of the first car = 14000 kg.
Initial speed of first car = 8.1 m/s.
Final speed of the two cars = 3.6 m/s.
From law of conservation of momentum; we can write:
Total initial momentum = total final momentum
14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
Hence, the mass of the second car is 17500 kg.
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What is 1 radian approximately in degrees?
Answer:
180 LUV:)
Explanation:
Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a
building. What is the kinetic energy at the bottom?
Answer:
972 J
Explanation:
At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:
GPE = mass×gravity×heigth
GPE = 2.2×9.8×45.08 ≈ 972
For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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A camera takes a properly exposed photo at f/5.6 and 1/250 s. What shutter speed should be used if the lens is changed to f/4.0?
a.1/65 s
b.1/125 s
c.1/250 s
d.1/500 s
e.1/1000 s
The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.
The relationship between the shutter speed and the focal ratio is an inverse relationship.
Ratio of Areas = 5.6²/4²
A1/A2 = 1.96 = 2Since; T1A1 = T2A2
T2 = T1(A1/A2)T2 = 250 × 2 = 500 secondsIn essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.
Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.
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a=100m/s2
Vi=10m/s
T=10sec
Vf?
Answer:
v=u+at
v= 10+100×10
v = 10+1000
v = 1010 m/s
The passage of a wave through a medium is...?
Answer:
Transmission
Explanation:
It’s the passage of electromagnetic waves through a medium.
The passage of a wave through a medium is known as transmission.
What is the transmission of a wave?The transmission of waves may be characterized as a wave traveling through a medium or into a new medium. Not all waves transmit through the same materials. It significantly depends on the type of wave, the type of material, and even the frequency of the wave.
Transmission is generally the passage of electromagnetic radiation through a medium. While Reflection is the process by which electromagnetic radiation is returned either at the boundary between two media (surface reflection) or at the interior of a medium (volume reflection).
Waves can go through things. Waves go through water, light can pass through the glass, and sound can pass through walls. The material the wave passes through is called a medium. A single disturbance passing through a medium is called a pulse.
Therefore, the passage of a wave through a medium is known as transmission.
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To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary, in what instance does a person become unhealthy
The axis of the earth is
Tilted about 23.5 degrees
Vertical
Vertical
Answer:
The axis of rotation of the Earth is tilted at an angle of 23.5 degrees away from vertical, perpendicular to the plane of our planet's orbit around the sun. The tilt of the Earth's axis is important, in that it governs the warming strength of the sun's energy.
Explanation:
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the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
2. If a vehicle with four wheels is traveling in a linear path through soft dirt, how many and which tires would you expect to leave tire track impressions
Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.
The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.
This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.
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Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=−kq, where k is a constant. Find the value of the constant k.
The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.
Thus, the relation for Kirchhoff's loop rule can be expressed as:
[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]
We all know that the current in the nonconstant charge flow can be written as:
[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]
∴
Replacing the current (I) into Kirchhoff's loop rule, we have:
[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]
From the given question, when the switch is closed
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]
Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.
∴
By equating both equations (1) and (2);
[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
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Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:
The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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7. A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then A) what is the maximum spring potential energy of the simple harmonic motion
At the equilibrium position of the spring, the mass has kinetic energy
1/2 (1.00 kg) (0.0400 m/s)² = 0.000800 J = 0.800 mJ = 800 µJ
and when the spring is maximally compressed, this kinetic energy is converted into potential energy.
The maximum spring potential energy of the simple harmonic motion is approximately 7.48 millijoules.
To find the maximum spring potential energy (Us) in simple harmonic motion, we need to determine the maximum displacement (A) of the mass from its equilibrium position.
In simple harmonic motion, the maximum displacement is equal to the amplitude of the oscillation.
The formula for maximum displacement (A) is given by:
A = v0 / ω
where v0 is the initial velocity (4.00 cm/s) and ω is the angular frequency of the oscillation.
The angular frequency (ω) is given by:
ω =[tex]\sqrt{(k / m)[/tex]
where k is the spring constant (9.00 N/m) and m is the mass (1.00 kg).
Substitute the values:
ω = [tex]\sqrt{(9.00 N/m / 1.00 kg)} = \sqrt{9.00 rad/s[/tex] ≈ 3.00 rad/s
Now, calculate the maximum displacement:
A = 4.00 cm/s / 3.00 rad/s ≈ 1.33 cm
Next, calculate the maximum spring potential energy:
Us =[tex](1/2) * k * A^2[/tex]
Us = (1/2) * 9.00 N/m * [tex](1.33 cm)^2[/tex]≈ 7.48 mJ
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--The compleete Question is, A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then what is the maximum spring potential energy of the simple harmonic motion--
How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is
∑ F = -f = ma = (4.00 kg) a
Assuming constant acceleration a , the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)
Solve for the acceleration :
a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-f = (4.00 kg) (-14.8 m/s²)
f ≈ 59.4 N
and was directed opposite the block's motion.
A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........