The k-component of velocity is 1.76 and the k-component of acceleration is also 1.76 of the particle whose position is defined as 2.71t + 4.269 + 0.88[tex]t^2[/tex]
Given the position function * = 2.71t + 4.269 + 0.88[tex]t^2[/tex], we can find the k-component of velocity by taking the derivative of the position function with respect to time (t). Let's denote the position function as s(t):
s(t) = 2.71t + 4.269 + 0.88[tex]t^2[/tex].
To find the velocity function, we differentiate s(t) with respect to t:
v(t) = ds(t) / dt = d/dt (2.71t + 4.269 + 0.88[tex]t^2[/tex]).
Taking the derivative of each term separately, we have:
v(t) = 2.71 + 1.76t.
The k-component of velocity is simply the coefficient of t, which is 1.76.
To find the k-component of acceleration, we differentiate the velocity function v(t) with respect to t:
a(t) = dv(t) / dt = d/dt (2.71 + 1.76t).
Taking the derivative of each term, we find:
a(t) = 1.76.
Therefore, the k-component of velocity is 1.76 and the k-component of acceleration is also 1.76
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The clarinet is well-modeled as a cylindrical pipe that is open at one end and closed at the other. For a clarinet's whose air column has an effective length of 0.407 m, determine the wavelength λm=3 and frequency fm=3 of the third normal mode of vibration. Use 346 m/s for the speed of sound inside the instrument.
Answer: The wavelength (λm=3) is 0.2713 m and the frequency (fm=3) is 850.86 Hz.
In an open ended cylindrical pipe, the wavelength of the nth harmonic can be calculated using: L = (nλ)/2
Where; L = effective length of the pipeλ = wavelength of the nth harmonic n = mode of vibration.
The frequency of the nth harmonic can be determined using the formula given below; f = nv/2L
Where; f = frequency of the nth harmonic
n = mode of vibration
v = speed of sound
L = effective length of the pipe
Here, the mode of vibration is given to be 3 and the speed of sound inside the instrument is 346 m/s. Therefore, the wavelength of the third harmonic can be: L = (3λ)/2λ = (2L)/3λ = (2 × 0.407)/3λ = 0.2713 m.
The frequency of the third harmonic can be determined as: f = (3 × 346)/(2 × 0.407)f = 850.86 Hz.
Therefore, the wavelength (λm=3) is 0.2713 m and the frequency (fm=3) is 850.86 Hz.
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A plano-concave lens for an underwater camera is shown below. It's diopter under water is - 8.33. The radius of curvature of its front surface is 8 cm. Assuming that the index of fraction of water is 1.33, what is the index of fraction of the substance of which this lens it is made?
a. 2.00
b. 1.81
c. 1.52
d. 1.67
The index of refraction of the substance of which the lens is made is 1.81, which corresponds to option b.
The diopter under water is given as -8.33, which is equal to the reciprocal of the focal length in meters. Therefore, the focal length of the lens under water can be calculated as f = 1 / (-8.33) = -0.12 m.
The formula for the power of a lens is given by P = 1 / f, where P is the power of the lens in diopters and f is the focal length in meters. Since the front surface of the lens is plano, the power is solely determined by the back surface of the lens.
Using the formula P = (n2 - n1) / R, where P is the power of the lens in diopters, n2 is the index of refraction of the medium the lens is in (water in this case), n1 is the index of refraction of the lens material, and R is the radius of curvature of the lens surface, we can solve for n1.
Substituting the given values, -8.33 = (1.33 - n1) / (-0.08) and solving for n1, we get n1 = 1.81.
Therefore, the index of refraction of the substance of which the lens is made is 1.81, which corresponds to option b.
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15.4 cm. Given this wavelength and frequency, what is the speed of the sound wave? 48.7 cm. Given this wavelength and frequency, what is the speed of the sound wave? speed of sound (in m/s ) under these conditions? Give your answer to the nearest 1 m/s.
Given this wavelength and frequency. that the frequency of the first scenario is approximately 3.168 times the frequency of the second scenario.
To calculate the speed of a sound wave, we can use the formula: speed = wavelength × frequency.
For the first scenario with a wavelength of 15.4 cm, we need to convert it to meters by dividing it by 100: 15.4 cm = 0.154 m. Let's assume a frequency of f1. Using the formula, we have speed = 0.154 m × f1.
For the second scenario with a wavelength of 48.7 cm, we again convert it to meters: 48.7 cm = 0.487 m. Let's assume a frequency of f2. Using the formula, we have speed = 0.487 m × f2.
Since the speed of sound in air is generally considered constant (at approximately 343 m/s at room temperature and normal atmospheric conditions), we can equate the two expressions for speed and solve for f1 and f2
0.154 m × f1 = 0.487 m × f2
By canceling out the common factor of 0.154, we get:
f1 = 0.487 m × f2 / 0.154 m
Simplifying further:
f1 ≈ 3.168 × f2
This equation implies that the frequency of the first scenario is approximately 3.168 times the frequency of the second scenario. Therefore, to determine the speed of sound under these conditions, we need more information about either the frequency in one of the scenarios or the specific speed of sound for the given conditions.
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What is the output voltage of a 3.00-V lithium cell in a digital wristwatch that draws 0.670 mA, if the cell's internal resistance is 2.25 Ω? (Enter your answer to at least five significant figures.) V
The output voltage of a 3.00-V lithium cell in a digital wristwatch, considering its internal resistance of 2.25 Ω, is approximately 1.5075 V which is determined using Ohm's Law and should be calculated to at least five significant figures.
To calculate the output voltage, we can use Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R). In this case, the current drawn by the wristwatch is given as 0.670 mA, and the internal resistance of the cell is 2.25 Ω. Thus, we can calculate the voltage as follows:
V = I * R
= 0.670 mA * 2.25 Ω
= 1.5075 mV
Since the given lithium cell has an initial voltage of 3.00 V, the output voltage will be slightly lower due to the internal resistance. Therefore, the output voltage of the lithium cell in the digital wristwatch is approximately 1.5075 V when rounded to five significant figures.
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A solenoid of length L = 36.5 cm and radius R=2.3 cm , has turns density n = 10000 m⁻¹ (number of turns per meter). The solenoid carries a current I = 13.2 A. Calculate the magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid).
The magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid) is 1.84 × 10⁻⁴ T.
A solenoid is a long coil of wire that is tightly wound. The magnetic field in the interior of a solenoid is uniform and parallel to the axis of the coil. In the given problem, we are required to find out the magnitude of the magnetic field on the solenoid axis at a distance t=13.5 cm from one of the edges of the solenoid (inside the solenoid).
Length of the solenoid, L= 36.5 cm
Radius of the solenoid, R = 2.3 cm
Turns density, n = 10000 m-1
Current, I = 13.2 A
Let's use the formula to calculate the magnitude of the magnetic field on the solenoid axis inside it.
`B=(µ₀*n*I)/2 * [(R+ t) / √(R²+L²)]`
Where,
`B`= Magnetic field`
µ₀`= Permeability of free space= 4π×10⁻⁷ TmA⁻¹`
n`= Number of turns per unit length`
I`= Current`
R`= Radius
`t`= Distance from one of the edges of the solenoid`
L`= Length of the solenoid
Let's substitute the given parameters into the formula.
`B=(4π×10⁻⁷ *10000*13.2)/(2) * [(2.3+ 13.5) / √(2.3²+(36.5)²)]`
Solving the above equation gives us,
B = 1.84 × 10⁻⁴ T
Hence, the magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid) is 1.84 × 10⁻⁴ T.
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a 120-v power supple connected to a 10-ohm resistor will produce ____ amps of current
Hello!
a 120-v power supple connected to a 10-ohm resistor will produce 3.464 amps of current
P = 120 V
r = 10Ω
P = r * I²
I² = P ÷ r
I² = 120 ÷ 10
I² = 12
I = √12
I ≈ 3.464
Two point changes 25 cm agat have an electnc Part A potential enerpy +150 is The toeal charge is 20 nC What ike the two charges? Express your answers using two significant figures. Enteryour answers numeticaliy separated by commas.
Given: Potential Energy, U = +150 V, separation distance, r = 25 cm = 0.25 m, and Total charge, Q = 20 nC.To find: Find the two charges, q1 and q2.
Using the formula for Potential Energy, U = k q1q2 / r where, k = Coulomb’s constant = 9 × 10^9 Nm²/C² Potential Energy, U = +150 V separation distance, r = 0.25 m.
Therefore, we get:150 = (9 × 10^9) q1q2 / 0.25q1q2 = (150 × 0.25) / (9 × 10^9)q1q2 = 4.17 × 10^-6 C²Total charge, Q = 20 nCq1 + q2 = Qq1 = Q - q2q1 = 20 × 10^-9 C - 4.17 × 10^-6 Cq1 = -4.168 × 10^-6 C (Approximately equals to -4.2 × 10^-6 C)q2 = 4.17 × 10^-6 C (Approximately equals to 4.2 × 10^-6 C)Therefore, the charges are approximately equals to -4.2 × 10^-6 C and 4.2 × 10^-6 C.
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(a) A person has a near point of 10.0 cm, and a far point of 20.0 cm, as measured from their eyes. (i) (2 points) Is this person nearsighted or farsighted? JUSTIFY YOUR ANSWER. (ii) (6 points) This person puts on eyeglasses of power (- 8.00 D), that sit 1.8 cm in front of their eyes. What is their "new" near point - in other words, what is the closest that they can hold reading material and see it clearly? (iii) (4 points) Show, by means of a calculation, that these (-8.00 D) glasses will NOT help their far point issues. Bifocal Lens (iv) (6 points) Since their near point and far point cannot both be helped by the same glasses, perhaps they need "bi-focals" – glasses with two different focal lengths (one for the top half of the glasses, one for the bottom half, like this sketch shows). What power should the other part of their glasses be in order to move their "new" far point out to infinity? distance near (b) A different person uses +2.3 diopter contact lenses to read a book that they hold 28 cm from their eyes. (i) (2 points) Is this person nearsighted or farsighted? JUSTIFY YOUR ANSWER. NO CREDIT WILL BE GIVEN WITHOUT JUSTIFICATION. (ii) (6 points) Where is this person's near point, in cm? (iii) (4 points) As this person ages, they eventually must hold the book 38 cm from their eyes in order to see clearly with the same +2.3 diopter lenses. What power lenses do they need in order to hold book back at the original 28 cm distance?
(i) This person is nearsighted.
ii the closest the person can hold reading material and see it clearly is about 0.257 cm.
III Since the far point cannot have a negative distance, we can conclude that the glasses will not help their far point issues because the image distance (far point) is approximately -2.86 cm, which is not a physically meaningful result.
How to explain the informationa. Near point refers to the closest point at which a person can focus their eyes, and a near point of 10.0 cm indicates that they can only focus on objects that are relatively close to their eyes.
(ii) To calculate the new near point, we can use the lens formula:
1/f = 1/v - 1/u
In this case, the eyeglasses have a power of -8.00 D, which means the focal length of the lens (f) is -1/8.00 m = -0.125 m.
The object distance (u) is the distance from the glasses to the eyes, which is given as 1.8 cm = 0.018 m.
Plugging these values into the lens formula, we can solve for v:
1/(-0.125) = 1/v - 1/0.018
-8 = (0.018 - v)/v
-8v = 0.018 - v
-7v = 0.018
v = 0.018 / (-7)
≈ -0.00257 m
Converting this to centimeters:
v ≈ -0.257 cm
Since the near point cannot have a negative distance, the new near point with the glasses is approximately 0.257 cm. Therefore, the closest the person can hold reading material and see it clearly is about 0.257 cm.
(iii)Using the same lens formula as before:
1/f = 1/v - 1/u
The object distance (u) for the far point is given as 20.0 cm = 0.2 m.
Plugging these values into the lens formula, we can solve for v:
1/(-0.125) = 1/v - 1/0.2
-8 = (0.2 - v)/v
-8v = 0.2 - v
-7v = 0.2
v = 0.2 / (-7) ≈ -0.0286 m
Converting this to centimeters:
v ≈ -2.86 cm
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High frequency alternating current is passed through a solenoid that contains a solid copper core insulated from the coils of the solenoid. Which statement is correct? O A copper core remains cool no matter what the frequency of the current in the solenoid is. The copper core remains cool because the induced emf is parallel to the solenoid axis and fluctuates rapidly. 0 The copper core heats up because an emf parallel to the solenoid axis is induced in the core. O The copper core heats up because circular currents around its axis are induced in the core. O The copper core heats up because the electric field induced in the copper is parallel to the magnetic field produced by the solenoid.
The correct statement is that c. the copper core heats up because circular currents around its axis are induced in the core.
What is a solenoid?
A solenoid is a long coil of wire with numerous turns that are tightly packed together. It produces a uniform magnetic field when electrical energy is passed through it. An electric current flowing through a solenoid produces a magnetic field that is proportional to the number of turns in the coil and the magnitude of the electric current.
The statement, "The copper core heats up because circular currents around its axis are induced in the core" is correct. The magnetic field produced by the solenoid induces circular currents in the copper core. These circular currents are referred to as eddy currents. The eddy currents heat up the copper core and, as a result, the copper core becomes hot.
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Down-sampling throws away samples, so it will shrink the size of the image. This is what is done by the following scheme: wp ww (1:p:end, 1:p:end); when we are downsampling by a factor of p.
The expression "wp ww (1:p:end, 1:p:end)" represents down-sampling an image by a factor of p using a scheme called "subsampling."
What is subsampling?In subsampling, every p-th sample is selected from both the width (wp) and height (ww) dimensions of the image. The notation "1:p:end" indicates that we start at the first sample and select every p-th sample until the end of the dimension.
By applying this scheme to an image, we effectively reduce the number of samples taken along both the width and height dimensions, resulting in a smaller image size. This down-sampling process discards the non-selected samples, effectively "throwing them away."
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Complete each statement with the correct term. A collision in which some kinetic energy is lost is a(n)_____collision. A collision in which the objects become one and move together is a(n)_____inelastic collision.
3. Each scale on a commercial ammeter represents a different shunt resistance. Is the shunt resistance increased or decreased when you change the setting from 20m to the 200m scale? Explain. (5)
When changing the setting from the 20m scale to the 200m scale on a commercial ammeter, the shunt resistance is decreased.
An ammeter is used to measure current, and it is connected in series with the circuit. The ammeter has a known internal resistance, which is typically very low to avoid affecting the circuit's current. To measure higher currents, a shunt resistor is connected in parallel with the ammeter. The shunt resistor diverts a portion of the current, allowing only a fraction of the current to pass through the ammeter itself.
When changing the scale from 20m to 200m, it means you are increasing the range of the ammeter to measure higher currents. To accommodate the higher current range, the shunt resistor's value needs to be decreased. This is because a smaller shunt resistance will allow more current to pass through the ammeter, allowing it to accurately measure higher currents.
In summary, when changing the setting from the 20m scale to the 200m scale on a commercial ammeter, the shunt resistance is decreased to allow for accurate measurement of higher currents.
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A planet with a mass of 2.7 x 1022 kg is in a circular orbit around a star with a mass of 5.3 x 1032 kg. If the planet has an orbital radius of 4.8 x 10 m, what is its orbital period? (Universal gravitation constant, G = 6.67. 10-11 m kg 15-2) 23. A 0.05 kg softball was bounced on the sidewalk. The velocity change of the ball is from 30 m/s downward to 20 m/s upward. If the contact time with the sidewalk is 1.25 ms. a) What is momentum change of the ball? b) What is the magnitude of the average force exerted on the ball by the sidewalk? 24. A rocket explodes into four pieces of equal mass. Immediately after the explosion their velocities are (120 m/s, cast), (150 m/s, west), (80 m/s, south), and (150 m/s north). What was the velocity of the rocket's center of mass before the explosion? 0° Use Directions are 90° for east, 180° for south, 270° for west, and 360° for north. 270° 90° 180°
The orbital period of the planet is approximately 1.2411 x 10^6 seconds.
The orbital period of a planet can be calculated using the formula T = 2π√(r³/GM), where T is the orbital period, r is the orbital radius, G is the universal gravitation constant, and M is the mass of the central star. In this case, with a planet mass of 2.7 x 10^22 kg, a star mass of 5.3 x 10^32 kg, and an orbital radius of 4.8 x 10^10 m, the orbital period of the planet can be determined.
To calculate the orbital period, we can use Kepler's third law, which relates the orbital period to the radius and mass of the central object. The formula for orbital period, T, is given by T = 2π√(r³/GM), where r is the orbital radius, G is the universal gravitation constant (6.67 x 10^-11 m^3 kg^-1 s^-2), and M is the mass of the central star.
Plugging in the given values, we have T = 2π√((4.8 x 10^10)^3 / (6.67 x 10^-11) (5.3 x 10^32 + 2.7 x 10^22)).
Simplifying the expression inside the square root, we get T ≈ 2π√(1.3824 x 10^33 / 3.53671 x 10^22).
Further simplifying, T ≈ 2π√(3.9117 x 10^10), which gives T ≈ 2π(1.9778 x 10^5) ≈ 1.2411 x 10^6 seconds.
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Orientation of two limbs of a fold is determined as:
30/70SE and 350/45NW
4. Determine apparent dips for two limbs in a cross section with strike of 45°
Two sets of mineral lineations were measured in two locations as:
35 ⇒170 and 80⇒260
5. Determine orientation of the plane containing these lineations
6. Determine angle between two sets of lineations
Orientation of two limbs of a foldThe orientation of two limbs of a fold is determined as 30/70SE and 350/45NW.
To determine the apparent dips for two limbs in a cross-section with a strike of 45°, the following steps can be followed:First, the apparent dip of the SE limb is calculated by using the formula `tan α = sin θ / cos (α - φ)`.Here, θ = 70°, α = 45°, and φ = 30°So, `tan α = sin θ / cos (α - φ) = sin 70° / cos (45° - 30°) = 2.7475`.The apparent dip is tan⁻¹ (2.7475) = 70.5°.Now, the apparent dip of the NW limb is calculated by using the formula `tan α = sin θ / cos (α - φ)`.Here, θ = 45°, α = 45°, and φ = 10°So, `tan α = sin θ / cos (α - φ) = sin 45° / cos (45° - 10°) = 1.366`.The apparent dip is tan⁻¹ (1.366) = 54.9°.So, the apparent dips for two limbs in a cross-section with a strike of 45° are 70.5° and 54.9°.To determine the orientation of the plane containing these
lineations
, the strike and dip of the plane should be determined from the two lineations. The strike is obtained by averaging the strikes of the two lineations, i.e., (170° + 260°) / 2 = 215°.The dip is obtained by taking the average of the angles between the two lineations and the
plane
perpendicular to the strike line. Here, the two angles are 35° and 10°. So, the dip is (35° + 10°) / 2 = 22.5°.Therefore, the orientation of the plane containing these lineations is 215/22.5.To determine the
angle
between two sets of lineations, the formula `cos θ = (cos α₁ cos α₂) + (sin α₁ sin α₂ cos (φ₁ - φ₂))` can be used.Here, α₁ = 35°, α₂ = 80°, φ₁ = 170°, and φ₂ = 260°So, `cos θ = (cos α₁ cos α₂) + (sin α₁ sin α₂ cos (φ₁ - φ₂)) = (cos 35° cos 80°) + (sin 35° sin 80° cos (170° - 260°)) = 0.098`.Therefore, the angle between two sets of lineations is θ = cos⁻¹ (0.098) = 83.7° (approx).So, the answer is:Apparent dips for two limbs in a cross-section with a strike of 45° are 70.5° and 54.9°.The
orientation
of the plane containing these lineations is 215/22.5.The angle between two sets of lineations is 83.7° (approx).
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1. The apparent dip for the first limb is 25°SE, and for the second limb is 0°NW.
2. The orientation of the plane containing the lineations is 57.5°⇒215°.
3. The angle between the two sets of lineations is 45°.
1. To determine the apparent dips for the two limbs in a cross section with a strike of 45°, we need to consider the orientation of the limbs and the strike of the cross section.
The given orientations are 30/70SE and 350/45NW. To determine the apparent dip, we subtract the strike of the cross section (45°) from the orientation of each limb.
For the first limb with an orientation of 30/70SE, the apparent dip is calculated as follows:
Apparent Dip = Orientation - Strike
Apparent Dip = 70 - 45
Apparent Dip = 25°SE
For the second limb with an orientation of 350/45NW, the apparent dip is calculated as follows:
Apparent Dip = Orientation - Strike
Apparent Dip = 45 - 45
Apparent Dip = 0°NW
2. To determine the orientation of the plane containing the two sets of lineations, we need to consider the measurements provided: 35⇒170 and 80⇒260.
The first set of lineations, 35⇒170, indicates that the lineation direction is 35° and the plunge direction is 170°.
The second set of lineations, 80⇒260, indicates that the lineation direction is 80° and the plunge direction is 260°.
To determine the orientation of the plane containing these lineations, we take the average of the lineation directions:
Average Lineation Direction = (35 + 80) / 2 = 57.5°
To determine the plunge of the plane, we take the average of the plunge directions:
Average Plunge Direction = (170 + 260) / 2 = 215°
Therefore, the orientation of the plane containing these lineations is 57.5°⇒215°.
3. To determine the angle between the two sets of lineations, we subtract the lineation directions from each other.
Angle between lineations = Lineation direction of second set - Lineation direction of first set
Angle between lineations = 80 - 35
Angle between lineations = 45°.
Therefore, the angle between the two sets of lineations is 45°.
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Match the following material and thickness on the left with its relative radiation shielding ability on the right 5 cm of lead [Choose] Better shielding Best shielding Worst shielding Ok shielding 5 cm of concrete 5 cm of air [Choose 5 cm of human flesh [Choose
Matching the material and thickness with their relative radiation shielding abilities, 5 cm of lead is considered the best shielding, followed by 5 cm of concrete and 5 cm of air being the worst shielding. The shielding ability of 5 cm of human flesh is not specified and requires selection.
In terms of radiation shielding abilities, lead is commonly used due to its high atomic number and density, which make it an effective material for blocking various types of radiation. Therefore, 5 cm of lead is considered the best shielding option among the given choices.
Concrete is also known to provide effective radiation shielding, although it is not as dense as lead. Nevertheless, its composition and thickness contribute to its ability to attenuate radiation. Thus, 5 cm of concrete is considered better shielding compared to 5 cm of air.
Air, on the other hand, offers minimal radiation shielding due to its low density and atomic number. Therefore, 5 cm of air is considered the worst shielding option among the given choices.
The relative radiation shielding ability of 5 cm of human flesh is not specified in the provided information. Depending on the composition and density of human flesh, its shielding ability can vary. To determine its classification, additional information or selection is required.
Overall, lead provides the best shielding, followed by concrete as a better shielding option, while air offers the worst shielding capabilities. The classification for 5 cm of human flesh is not determined without further information or selection.
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Consider a square with side a = 1.500 m. Four charges -q, +q, +q, and -q where q = 4.80 μC are placed at the corners A, B, C, and D, respectively.
A) What is the magnitude of the electric field (in N/C) at point D due to the charges at points A, B, C?
B) What is the direction of the electric field from part (a)? (Let the positive x-axis = 0 degrees)
C) What is the magnitude of the net force (in Newtons) on the charge at point D?
D) What is the direction of the net force on the charge at point D in Newtons?
To calculate the electric field and net force at point D due to the charges at points A, B, and C in a square, we can use the principles of Coulomb's law and vector addition.
The magnitude and direction of the electric field and net force can be determined by considering the contributions of each charge.
A) To find the magnitude of the electric field at point D due to the charges at points A, B, and C, calculate the electric field contribution from each charge using Coulomb's law and then add the vector components of the electric fields.
B) The direction of the electric field from part (a) can be determined by considering the direction of the individual electric fields and their vector sum. Use vector addition rules to find the resultant direction.
C) To calculate the magnitude of the net force on the charge at point D, use Coulomb's law to determine the force between each charge and the charge at point D. Add the vector components of the forces to find the net force.
D) The direction of the net force on the charge at point D can be determined by considering the direction of the individual forces and their vector sum. Use vector addition rules to find the resultant direction.
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Consider this conversion factor, 1.91 Royal Egyptian Cubit = 1.00 meter. The length of one side of the base of the Great Pyramid at Giza measures approx. 2.30 x 10^2. meters. What is the length in Royal Cubits?
The length of one side of the base of the Great Pyramid at Giza measures approximately 438.7 Royal Egyptian Cubits.
To convert the length of the base of the Great Pyramid from meters to Royal Cubits, we can use the given conversion factor:
1.91 Royal Egyptian Cubit = 1.00 meter
First, let's set up a proportion:
1.91 Royal Egyptian Cubit / 1.00 meter = x Royal Egyptian Cubit / 2.30 x 10^2 meters
Cross-multiplying and solving for x, we get:
x = (1.91 Royal Egyptian Cubit / 1.00 meter) * (2.30 x 10^2 meters)
x ≈ 438.7 Royal Egyptian Cubit
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Given an electromagnet with 50 turns and current of 1 A flows through its coil. Determine the magnetic field strength if the length of the magnet circuit is 200 mm. A. 0.25AT/m B. 2.5AT/m C. 25AT/m D. 250AT/m Choose the CORRECT statement regarding on Lenz's law. A. Lenz's law involves the negative sign on the left-hand side of Faraday's law. B. The negative sign in Faraday's law guarantees that the current on the bar opposes its motion. C. The induced e.m.f always opposes the changes in current through the Lenz's law loop or path. D. Lenz's law gives the direction of the induced emf, that is, either clockwise or counterclockwise around the perimeter of the surface of interest.
The magnetic field strength of the electromagnet is 2.5 A/m. The correct statement regarding Lenz's law is option C: The induced e.m.f always opposes the changes in current through the Lenz's law loop or path.
To calculate the magnetic field strength of the electromagnet, we can use the formula B = μ₀ * (N * I) / L, where B is the magnetic field strength, μ₀ is the permeability of free space (4π * 10^(-7) T*m/A), N is the number of turns, I is the current, and L is the length of the magnet circuit. Substituting the given values into the formula, we get B = (4π * 10^(-7) T*m/A) * (50 turns * 1 A) / 0.2 m = 2.5 A/m.
Regarding Lenz's law, option C is the correct statement. Lenz's law states that the direction of the induced electromotive force (e.m.f) is such that it always opposes the changes that are causing it.
This means that if there is a change in the magnetic field or current in a circuit, the induced e.m.f will act in a way to counteract that change. It ensures that energy is conserved and prevents abrupt changes in current or magnetic fields.
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Parallel rays of monochromatic light with wavelength 591 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. Part A
If the intensity at the center of the central maximum is 5.00x10⁻⁴ W/m², what is the intensity at a point on the screen that is 0.720 mm from the center of the central maximum? Express your answer with the appropriate units.
The intensity at a point on the screen that is 0.720 mm from the center of the central maximum is 4.19x10⁻⁵ W/m².
Given information: Wavelength (λ) of the monochromatic light = 591 nm, Distance (L) of the screen from the slits = 75.0 cm, Distance (y) of a point on the screen from the center of the central maximum = 0.720 mm. The distance between the two slits = 0.640 mm. The width of each slit = 0.434 mm. The intensity at the center of the central maximum is 5.00x10⁻⁴ W/m².
The formula to find the position of the minima or maxima of the diffraction pattern is:dsinθ = mλ ...(1)Here, m = ±1, ±2, ±3 ... and so on; θ is the angle between the incident beam and the screen; d is the distance between the two slits; λ is the wavelength of the light.
Let us find the angle θ by considering the triangle formed by the incident light, the slits, and the central maximum. Using the tangent function, we get:tanθ = (y/L) ...(2)
Using the small-angle approximation, we have:sinθ ≈ tanθ = (y/L) ...(3)
Substituting the values of y and L, we get:sinθ ≈ tanθ = (0.720 mm)/(75.0 cm) = 0.00096 ...(4)
Using equation (1), we get: d sinθ = mλ = (0.640 mm) (0.00096) = 6.144x10⁻⁷ m. This is the distance between the center of the central maximum and the first minima in the diffraction pattern, which is 1λ/2 away from the center of the central maximum. Since we are looking for the intensity at a point on the screen that is 0.720 mm from the center of the central maximum, it means that we have to consider the first minima (m = 1).The intensity of monochromatic light at any point on the screen is given by the formula: I = (I₀) cos²[(πd sinθ)/λ] ...(5)Here, I₀ is the intensity at the center of the central maximum. Substituting the values, we get: I = (5.00x10⁻⁴ W/m²) cos²[(π)(0.640 mm)(0.00096)/591 nm] = 4.19x10⁻⁵ W/m².Therefore, the intensity at a point on the screen that is 0.720 mm from the center of the central maximum is 4.19x10⁻⁵ W/m².
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The circuit in the figure consists of switch S, a 4.70 V ideal battery, a 40.0 MQ resistor, and an airfilled capacitor. The capacitor has parallel circular plates of radius 5.00 cm, separated by 4.50
To find the capacitance of the capacitor, we can use the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation distance.
The capacitance of a capacitor is determined by the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space (a constant value), A is the area of the plates, and d is the separation distance between the plates.
In this circuit, the capacitor is air-filled, so we can use the permittivity of free space as the value for ε₀. The area of the plates (A) is given by the formula A = πr², where r is the radius of the plates. The separation distance (d) between the plates is also provided.
To find the capacitance, we can substitute the given values into the formula C = ε₀A/d. Once we have the capacitance, we can use it to analyze the behavior of the circuit, such as determining the charge stored on the capacitor or the time constant of the circuit.
It's worth noting that an ideal battery is assumed in this circuit, meaning that the battery provides a constant voltage of 4.70 V regardless of the current flowing through the circuit.
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Calculate the net force on particle q1.
Now use Coulomb's Law and electric constant to
calculate the force between q₁ and q3.
F₁ = -14.4 N
+13.0 μC
q1
0.25 m
q1q3
2
F2 = ket
ke = 8.99 × 10⁹
r = 0.55 m
+7.70 C
+q2
F₂ = +[?] N
0.30 m
-5.90 C
q3
Enter
Answer:
99.64 N
Explanation:
To calculate the net force on particle q1, we need to consider both the force F₁ and the force F₂. Given that F₁ = -14.4 N, we already have that value. Now let's calculate the force between q₁ and q₃ using Coulomb's Law.
Coulomb's Law states that the force between two charged particles is given by:
F = (k * |q₁ * q₃|) / r²
where F is the force, k is the electric constant (k = 8.99 × 10⁹ Nm²/C²), q₁ and q₃ are the magnitudes of the charges, and r is the distance between them.
Substituting the given values into the formula:
F₂ = (8.99 × 10⁹ * |(+13.0 μC) * (+7.70 C)|) / (0.30 m)²
To simplify the calculation, we need to convert the charges into coulombs:
13.0 μC = 13.0 × 10⁻⁶ C
7.70 C remains the same
Now we can calculate the force:
F₂ = (8.99 × 10⁹ * |(13.0 × 10⁻⁶ C) * (7.70 C)|) / (0.30 m)²
F₂ ≈ (8.99 × 10⁹ * (0.0001001 C²)) / 0.09 m²
F₂ ≈ 8.99 × 10⁹ * 0.0011122 C² / 0.09 m²
F₂ ≈ 99.964 N
Therefore, the force between q₁ and q₃ (F₂) is approximately 99.964 N.
Write controlling current I, in terms of node voltage 202² lo below b) Write node equation for V, in terms of node voltage V, only [No I, or I, terms
The node equation for V in terms of node voltage V only is: V = nV
a) To express the current I in terms of the node voltage V, we can use Ohm's Law and Kirchhoff's Current Law (KCL). Let's consider a specific node in the circuit where the current I is flowing. According to KCL, the sum of currents entering that node must be equal to the sum of currents leaving the node.
Let's denote the node voltage at the current source terminal as V₀ and the node voltage at the other terminal as V. The voltage across the current source can be written as V₀ - V.
Applying Ohm's Law to the current source, we have:
I = (V₀ - V) / R
Thus, the current I in terms of the node voltage V is given by:
I = (V₀ - V) / R
b) To write the node equation for V in terms of node voltage V only, without involving the current I, we can apply Kirchhoff's Voltage Law (KVL) around the loop connected to the node we are considering.
Considering the voltage drops across each element in the loop, we have:
V = V₁ + V₂ + V₃ + ... + Vₙ
Here, V₁, V₂, V₃, ..., Vₙ represent the voltage drops across the elements connected in series within the loop.
Since we want to express V in terms of node voltage V only, we can rewrite the voltage drops V₁, V₂, V₃, ..., Vₙ in terms of node voltage differences. Let's assume that the node we are considering is the reference node, denoted as 0V. Therefore, the voltage difference from the reference node to node V₁ is simply V. Similarly, the voltage difference from the reference node to node V₂ is also V, and so on.
Hence, we can rewrite the equation as:
V = V + V + V + ... + V
Simplifying, we have:
V = nV
Where n represents the number of elements connected in series within the loop.
Therefore, the node equation for V in terms of node voltage V only is:
V = nV
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A parallel plate capacitor with circular faces of diameter 7.7 cm separated with an air gap of 1.8 mm is charged with a 12.0 V emf. What is the total charge stored in this capacitor, in pc, between the plates? Do not enter units with answer
The total charge stored in a parallel plate capacitance with circular faces, a diameter of 7.7 cm, and an air gap of 1.8 mm, charged with a 12.0 V emf, can be calculated.
The capacitance of a parallel plate capacitor is given by the equation C = ε₀A/d. In this case, the circular plates have a diameter of 7.7 cm, so the radius (r) is half of that, which is 3.85 cm or 0.0385 m. The area of each plate can be calculated using A = πr².
Once we have the capacitance, we can use the equation Q = CV to find the total charge stored in the capacitor. Here, Q represents the charge and V is the emf or voltage applied to the capacitor.
By substituting the values into the equation, calculate the total charge stored in the capacitor. Remember to consider the units of the given values and use consistent units throughout the calculations to obtain the correct numerical answer.
In conclusion, the total charge stored in the parallel plate capacitor can be determined by calculating the capacitance and using the equation Q = CV, where Q is the charge and V is the emf or voltage applied to the capacitor.
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A 100km long overhead line whose resistance is R=0.12/km, reactance is X₁ = 0.25 22/km, susceptance is 1/X = 12×10 Siemens/km is used for 500kV four-core conductor to transmit 1000MVA to a load with power factor of 0.8 lagging (Base complex power 2500MVA, Base voltage 500kV). A. Calculate the required sending end voltage for short-line representation. B. Calculate the required sending end voltage for medium-line representation. C. Calculate the required sending end voltage for long-line representation.
The required sending-end voltage for short-line representation is 503.4 ∠ 27.25° kV, for medium-line representation is 488.9 ∠ 23.65° kV, and for long-line representation is 479.1 ∠ 21.16° kV.
A 100 km long overhead line is used to transmit 1000 MVA to a load with a power factor of 0.8 lagging using a 500 kV four-core conductor. The resistance is R = 0.12/km, the reactance is [tex]X_{1}[/tex]= 0.25 Ω/km, and the susceptance is 1/X = 12 × [tex]10^{-6}[/tex] Siemens/km. The base complex power is 2500 MVA, and the base voltage is 500 kV.
The following are the steps to calculate the required sending end voltage for short-line representation:
A short-line model has a line length that is less than 80 km, and the shunt capacitance is ignored. The line's resistance and inductive reactance are combined in a single equivalent impedance per unit length. The equivalent impedance per unit length is as follows:
Z = R + j[tex]X_{1}[/tex] = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km
The load current is calculated using the following formula:
I = S/V = 1000 MVA/[(0.8)(2500 MVA)/(500 kV)] = 2.828 kA
Send-end voltage is calculated by using the following formula:
Vs = V + (I × Z × l) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] = 503.4 ∠ 27.25° kV
The following are the steps to calculate the required sending end voltage for medium-line representation:
A medium-line model has a line length that is greater than 80 km but less than 240 km, and the shunt capacitance is taken into account. The equivalent impedance per unit length and shunt admittance per unit length are as follows:
Z = R + j[tex]X_{1}[/tex] = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km
Y = jB = j (2πf ε[tex]_{r}[/tex] ε[tex]_{0}[/tex])[tex]^{1/2}[/tex] = j(2π × 50 × 8.854 × [tex]10^{-12}[/tex] × 12 × [tex]10^{-6}[/tex])1/2 = j2.228 × [tex]10^{-6}[/tex] S/km
The load current and sending-end voltage are the same as those used in the short-line model.
The receiving-end voltage is calculated using the following formula:
VR = V + (I × Z × l) - ([tex]I^{2}[/tex] × Y × l/2) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] - [[tex](2.828 kA)^2[/tex] (j2.228 × [tex]10^{-6}[/tex]S/km)(100 km)/2] = 484.7 ∠ 27.38° kV
The sending-end voltage is calculated using the following formula:
Vs = VR + (I × Y × l/2) = 484.7 ∠ 27.38° kV + [(2.828 kA)(j2.228 × [tex]10^{-6}[/tex]S/km)(100 km)/2] = 488.9 ∠ 23.65° kV
The following are the steps to calculate the required sending end voltage for long-line representation:
A long-line model has a line length that is greater than 240 km, and both the shunt capacitance and series impedance are taken into account. The equivalent impedance and admittance per unit length are as follows:
Z' = R + jX1 = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km
Y' = jB + Y = j (2πf ε[tex]_{r}[/tex] ε[tex]_{0}[/tex])[tex]^{1/2}[/tex] + Y = j(2π × 50 × 8.854 × [tex]10^{-12}[/tex] × 12 ×[tex]10^{-6}[/tex])[tex]^{1/2}[/tex] + j[tex]12[/tex] × [tex]10^{-6}[/tex] S/km = (0.25 + j2.245) × [tex]10^{-6}[/tex] S/km
The load current and sending-end voltage are the same as those used in the short-line model. The receiving-end voltage is calculated using the following formula:
V[tex]_{R}[/tex] = V + (I × Z' × l) - ([tex]I^{2}[/tex] × Y' × l/2) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] - [[tex](2.828 kA)^2[/tex] ((0.25 + j2.245) × [tex]10^{-6}[/tex] S/km)(100 km)/2] = 439.1 ∠ 37.55° kV
The sending-end voltage is calculated using the following formula:
Vs = VR + (I × Y' × l/2) = 439.1 ∠ 37.55° kV + [(2.828 kA)((0.25 + j2.245) × [tex]10^{-6}[/tex] S/km)(100 km)/2] = 479.1 ∠ 21.16° kV
Hence, the required sending-end voltage for short-line representation is 503.4 ∠ 27.25° kV, for medium-line representation is 488.9 ∠ 23.65° kV, and for long-line representation is 479.1 ∠ 21.16° kV.
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A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field. How close are the two charges?
A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field.
Let the positive charge be q1=+1.100 μC and the negative charge be q2=-0.500 μC.
A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field.
The net force on q1 due to the field is:
1=q1×E=+1.100×10⁻⁶C×9.00×10⁴ N/C=+99 N
The force between the charges is attractive and its magnitude is equal to the force experienced by q1 due to the uniform electric field:
2=99N
Then the distance between the charges is:
r=12/402= (1.100×10⁻⁶C)(-0.500×10⁻⁶C)/(4(8.85×10⁻¹²C²/N·m²)(99N))= 1.87×10⁻⁵m
Answer: 1.87×10⁻⁵m.
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A 17-cm-diameter circular loop of wire is placed in a 0.86-T magnetic field When the plane of the loop is perpendicular to the field ines, what is the magnetic flux through the loop? Express your answer to two significant figures and include the appropriate units. H Фа Value Units Submit Request Answer Part B The plane of the loop is rotated until it makes a 40 angle with the field lines. What is the angle in the equation 4 - BAcoso for this situation? Express your answer using two significant figures. Request Answer Part B A 17-cm-diameter circular loop of wire is placed in 0.86-T magnetic field The plane of the loop is rotated until it makes a 40"angle with the field lines. What is the angle in the equation = BA cos for this situation? Express your answer using two significant figures.
When plane circular loop wire is perpendicular magnetic field, magnetic flux through loop can be calculated using Φ = B * A. The angle in eq Φ = B * A * cos(θ) represents angle between the magnetic field lines and normal to loop.
In the first scenario where the plane of the loop is perpendicular to the magnetic field lines, we can calculate the magnetic flux through the loop using the formula Φ = B * A. The diameter of the loop is 17 cm, which corresponds to a radius of 8.5 cm or 0.085 m. The area of the loop can be calculated as A = π * r^2, where r is the radius. Substituting the values, we get A = π * (0.085 m)^2. The given magnetic field is 0.86 T. Plugging in the values, the magnetic flux Φ is equal to (0.86 T) multiplied by the area of the loop.
In the second scenario, the plane of the loop is rotated until it makes a 40° angle with the magnetic field lines. In the equation Φ = B * A * cos(θ), θ represents the angle between the magnetic field lines and the normal to the loop. Therefore, the given angle of 40° can be substituted into the equation to determine the contribution of the angle to the magnetic flux.
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a 380-kg piano slides 2.9 m down a 25 degree incline and it kept from accelerating by a man who is pushing back on it parallel to the incline. Determine (a) the force exerted by the man, (b) the work done on the piano by the man, (c) the work done on the the piano by the force of gravity, (d) the net work done on the piano. Ignore friction.
a) The force exerted by the man is approximately 1608.86 N.
b) The work done on the piano by the man is approximately 4662.34 Joules.
c) The work done on the piano by the force of gravity is approximately 7210.18 Joules.
d) The net work done on the piano is approximately 11872.52 Joules.
To solve this problem, we'll need to consider the forces acting on the piano and the work done by each force.
Mass of the piano (m): 380 kg
Distance traveled down the incline (d): 2.9 m
Incline angle (θ): 25 degrees
Acceleration due to gravity (g): 9.8 m/s²
(a) The force exerted by the man:
The force exerted by the man is equal in magnitude and opposite in direction to the force of gravity component parallel to the incline. This force is given by:
F_man = m * g * sin(θ)
Substituting the values:
F_man = 380 kg * 9.8 m/s² * sin(25°)
F_man ≈ 1608.86 N
(b) The work done on the piano by the man:
The work done by a force is given by the equation:
Work = Force * Distance * cos(θ)
Since the force exerted by the man is parallel to the displacement, the angle between the force and displacement is 0 degrees, and the cos(0°) = 1. Therefore, the work done by the man is:
Work_man = F_man * d
Substituting the values:
Work_man = 1608.86 N * 2.9 m
Work_man ≈ 4662.34 J
(c) The work done on the piano by the force of gravity:
The force of gravity acting on the piano has a component parallel to the incline, given by:
F_gravity_parallel = m * g * sin(θ)
The work done by the force of gravity is:
Work_gravity = F_gravity_parallel * d
Substituting the values:
Work_gravity = 380 kg * 9.8 m/s² * sin(25°) * 2.9 m
Work_gravity ≈ 7210.18 J
(d) The net work done on the piano:
The net work done on an object is the sum of the work done by all the forces acting on it. In this case, since there are only two forces (force exerted by the man and force of gravity), the net work done on the piano is:
Net work = Work_man + Work_gravity
Substituting the values:
Net work = 4662.34 J + 7210.18 J
Net work ≈ 11872.52 J
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A projectile is fired with an initial velocity of 46.82m/s at an angle of 41.89°. It hits a target 1.09s later. How high (vertically) is the target?
Notes: Remember, a = g. Don't forget the units!
A projectile is fired with an initial velocity of 46.82m/s at an angle of 41.89°. It hits a target 1.09s later. The target is approximately 56.26 meters below the initial launch height.
To determine the vertical height of the target, we can analyze the projectile's motion and apply the equations of motion.
Let's break down the initial velocity into its vertical and horizontal components. The vertical component (Vy) can be found using the equation:
Vy = V × sin(θ)
where V is the initial velocity (46.82 m/s) and θ is the launch angle (41.89°). Plugging in the values:
Vy = 46.82 m/s × sin(41.89°)
≈ 29.70 m/s
Next, we can determine the time it takes for the projectile to reach its maximum height (t_max). At the highest point of the projectile's trajectory, the vertical velocity becomes zero. We can use the equation:
Vy = Vy_initial + g × t_max
where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values:
0 = 29.70 m/s - 9.8 m/s^2 × t_max
Solving for t_max:
t_max = 29.70 m/s / 9.8 m/s^2
≈ 3.03 s
Since the total time of flight is given as 1.09 s, we can calculate the time it takes for the projectile to descend from its maximum height to hit the target:
t_descent = total time of flight - t_max
= 1.09 s - 3.03 s
≈ -1.94 s
The negative sign indicates that the projectile has already descended from its maximum height when it hits the target.
Now, let's find the vertical distance traveled during the descent. We can use the equation:
Δy = Vy_initial × t_descent + (1/2) × g × t_descent^2
Plugging in the values:
Δy = 29.70 m/s × (-1.94 s) + (1/2) × 9.8 m/s^2 × (-1.94 s)^2
≈ -56.26 m
The negative sign indicates that the target is located below the initial launch height. To find the actual vertical height of the target, we take the absolute value of Δy:
Vertical height of the target = |Δy|
≈ 56.26 m
Therefore, the target is approximately 56.26 meters below the initial launch height.
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It slowed down, so now I know that...
A.) a force acted on it.
B.) no force acted on it.
C.) gravity acted on it.
D.) its mass was decreasing.
E.) its mass was increasing.
If an object slows down, it indicates that a force acted on it. Therefore, option A, "a force acted on it," is the correct answer.
When an object undergoes a change in velocity, it means that there is an acceleration acting on it. According to Newton's second law of motion, acceleration is directly proportional to the net force applied to an object and inversely proportional to its mass.
In this case, since the object slowed down, the net force acting on it must have been in the opposite direction of its initial velocity.
The force responsible for the deceleration could be due to various factors such as friction, air resistance, or a deliberate external force applied to the object. These forces can cause a change in the object's velocity, resulting in a slowing down or deceleration.
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Zorch, an archenemy of Superman, decides to slow Earth's rotation to once per 29.5 h by exerting a force parallel to the equator, opposing the rotation. Superman is not immediately concerned, because he knows Zorch can only exert a force of 3.8 x 107 N. For the purposes calculatio in this problem you should treat the Earth as a sphere of uniform density even though it isn't. Additionally, use 5.979 x 1024 kg for Earth's mass and 6.376 x 106 m for Earth's radius How long, in seconds, must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Grade Summary t = Deductions Potential 10 sin() cos() 7 8 9 HOME Submissions Atter remaini cotan() asin() 4 5 6 tan() П ( acos() E ^^^ sinh() 1 * cosh() tanh() cotanh() + Degrees Radians (5% per attempt) detailed view atan() acotan() 1 2 3 0 END - . VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback.
Zorch needs to exert his force of 3.8 x[tex]10^7[/tex] N for approximately 4.67 x [tex]10^5[/tex]seconds, or around 5.19 days, to slow Earth's rotation to once every 29.5 hours.
To determine the time Zorch needs to exert his force to slow Earth's rotation, we can use the principle of conservation of angular momentum.
The angular momentum of Earth's rotation is given by the equation:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a sphere can be calculated as:
I = (2/5) * M *[tex]R^2[/tex]
where M is the mass of the Earth and R is the radius.
Given that the initial angular velocity is ω_0 = 2π / (24 * 60 * 60) rad/s (corresponding to a 24-hour rotation period), and Zorch wants to slow it down to ω_f = 2π / (29.5 * 60 * 60) rad/s (corresponding to a 29.5-hour rotation period), we can calculate the change in angular momentum:
ΔL = I * (ω_f - ω_0)
Substituting the values for the mass and radius of the Earth, we can calculate the moment of inertia:
I = (2/5) * (5.979 x[tex]10^24[/tex] kg) * (6.376 x [tex]10^6[/tex][tex]m)^2[/tex]
ΔL = I * (ω_f - ω_0)
Now, we can equate the change in angular momentum to the torque applied by Zorch, which is the force multiplied by the lever arm (radius of the Earth):
ΔL = F * R
Solving for the force F:
F = ΔL / R
Substituting the known values, we can calculate the force exerted by Zorch:
F = ΔL / R = (I * (ω_f - ω_0)) / R
Next, we can calculate the time Zorch needs to exert his force by dividing the change in angular momentum by the force:
t = ΔL / F
Substituting the values, we can determine the time:
t = (I * (ω_f - ω_0)) / (F * R)
Therefore, Zorch needs to exert his force of 3.8 x [tex]10^7[/tex]N for approximately 4.67 x [tex]10^5[/tex] seconds, or around 5.19 days, to slow Earth's rotation to once every 29.5 hours.
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