a) The object's average velocity between t1 and t2 is simply 3
b) The object's instantaneous velocity at any given time is 3.
c) The object's average acceleration between t1 and t2 is 0
d) The object's instantaneous velocity when its acceleration is zero is also 3
How to find the object's average velocity between t1 and t2?a) To find the object's average velocity between t1 and t2, we can use the formula:
Average velocity = (Δx) / (Δt)
Given that the object's position as a function of time is given by x(t) = 3t - 0.541, we can find the displacement Δx between t1 and t2 by subtracting x(t1) from x(t2):
Δx = x(t2) - x(t1)
Using the given expression for x(t), we have:
Δx = [(3t2 - 0.541) - (3t1 - 0.541)]
= 3(t2 - t1)
The average velocity is then:
Average velocity = Δx / (Δt)
= [3(t2 - t1)] / (t2 - t1)
= 3
Therefore, the object's average velocity between t1 and t2 is simply 3 (with proper SI units).
How to find the object's instantaneous velocity at a specific time t?b) To find the object's instantaneous velocity at a specific time t, we can take the derivative of the position function x(t) with respect to time:
Instantaneous velocity = dx/dt
Given that x(t) = 3t - 0.541, the derivative of x(t) with respect to t is:
dx/dt = 3
Therefore, the object's instantaneous velocity at any given time is 3 (with proper SI units).
How to find the object's average acceleration between t1 and t2?c) To find the object's average acceleration between t1 and t2, we can again use the formula:
Average acceleration = (Δv) / (Δt)
Here, Δv represents the change in velocity, which is given by the difference between the instantaneous velocities at t1 and t2:
Δv = v(t2) - v(t1)
Since the object's instantaneous velocity is constant and equal to 3 (as calculated in part b), we have:
Δv = (3 - 3) = 0
The average acceleration is then:
Average acceleration = Δv / (Δt)
= 0 / (t2 - t1)
= 0
Therefore, the object's average acceleration between t1 and t2 is 0 (with proper SI units).
How to find if the object's acceleration is zero?d) If the object's acceleration is zero, it means that its velocity is constant. In this case, the object's instantaneous velocity will be the same as its average velocity.
Since the average velocity was previously calculated as 3 (with proper SI units), the object's instantaneous velocity when its acceleration is zero is also 3 (with proper SI units).
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When the distance from object to a thin convex lens is less than the focal length, the image will be QS:- Optical fibers are a modern technology used to transfer information. The main optical phenomenon that is used in work of optical fiber is Q9:- Given the wave function of magnetic component (in $1 units) for a sodium vellew light wave B(z,t)=B
0
sin2π(1.7×10
6
z−5.1×10
13
t). The energy for this photon of light (in electrun volis) is liquid-diamond (n
1
=1.37.n
1
=2.418) interface is index of the prism if the desiation angle eqaal 11
∘
The main optical phenomenon used in optical fibers is total internal reflection.
Given the wave function of the magnetic component (in $1 units) for a sodium yellow light wave B(z, t) = B₀ sin(2π(1.7×10⁶z - 5.1×10¹³t)). The energy for this photon of light (in electron volts) is liquid-diamond (n₁ = 1.37, n₂ = 2.418) interface is the index of the prism if the deviation angle equals 11°.
To determine the index of the prism, we can use Snell's law, which states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the indices of refraction (n₁ and n₂) of the two media:
n₁ sin(θ₁) = n₂ sin(θ₂)
In this case, the light is incident from a medium with an index of 1.37 (liquid) onto a medium with an index of 2.418 (diamond). Let's assume that the angle of incidence (θ₁) is equal to the deviation angle (θ) of 11°.
n₁ sin(θ) = n₂ sin(θ₂)
Since we are given the indices of refraction (n₁ = 1.37, n₂ = 2.418) and the deviation angle (θ = 11°), we can solve for θ₂:
sin(θ₂) = (n₁ / n₂) sin(θ)
sin(θ₂) = (1.37 / 2.418) sin(11°)
sin(θ₂) = 0.5659
Now, to determine the index of the prism, we need to calculate the angle of refraction (θ₂) and then use Snell's law again:
n₂ = (n₁ / sin(θ₁)) sin(θ₂)
n₂ = (1.37 / sin(11°)) sin⁻¹(0.5659)
n₂ ≈ 1.829
Therefore, the index of the prism is approximately 1.829.
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A water turbine is to generate 3.75 MW at 250 rpm under a head of 12 m from a hydro dam. A new same geometrical turbine design is to be fabricated to generate 2.25 MW under 7.5m head for another hydro dam. Determine the following:
a) the new turbine operation speed
b) the diameter ratio of the new turbine to the old turbine
c) the specific speed for both turbines.
a) The new turbine operation speed is approximately 167 rpm.
b) The diameter ratio of the new turbine to the old turbine is approximately 0.71.
c) The specific speed for both turbines is approximately 84.
To determine the new turbine operation speed, we can use the concept of specific speed (Ns). Specific speed is a dimensionless number that represents the rotational speed of a turbine relative to its size and the head under which it operates. The formula for specific speed is given by:
Ns = (N * √P) / H^0.75
where N is the rotational speed in RPM, P is the power output in kilowatts (kW), and H is the head in meters.
For the given information about the old turbine, we know it operates at 250 RPM and generates 3.75 MW (3,750 kW) under a head of 12 m. Plugging these values into the specific speed formula, we can calculate the specific speed for the old turbine as follows:
Ns_old = (250 * √3,750) / 12^0.75 ≈ 133.63
Now, for the new turbine, we are given that it needs to generate 2.25 MW (2,250 kW) under a head of 7.5 m. We need to determine its operation speed and the diameter ratio relative to the old turbine. Since the specific speed is a constant for turbines of the same geometry, we can set up the following equation:
Ns_old = N_new * (√P_new / P_old) * (H_old / H_new)^0.75
Substituting the known values:
133.63 = N_new * (√2,250 / 3,750) * (12 / 7.5)^0.75
Simplifying the equation and solving for N_new, we find:
N_new ≈ 167 RPM
To determine the diameter ratio (D_new / D_old), we can use the following relationship:
(D_new / D_old) = (N_old / N_new) * (√P_new / √P_old) * (H_old / H_new)^0.25
Substituting the known values:
(D_new / D_old) = (250 / 167) * (√2,250 / √3,750) * (12 / 7.5)^0.25
Simplifying the equation, we find:
(D_new / D_old) ≈ 0.71
Finally, the specific speed for both turbines can be calculated using the formula mentioned earlier. The specific speed is a constant, so it remains the same for both turbines:
Ns = (N * √P) / H^0.75
For the old turbine:
Ns_old = (250 * √3,750) / 12^0.75 ≈ 133.63
And for the new turbine:
Ns_new = (167 * √2,250) / 7.5^0.75 ≈ 133.63
Hence, the specific speed for both turbines is approximately 84.
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An elevator filled with passengers has a mass of 1583 kg.
(a)
The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.75 s. Calculate the tension in the cable (in N) supporting the elevator.
(b) The elevator continues upward at constant velocity for 8.72 s. What is the tension in the cable
(in N) during this time?
(c)
The elevator decelerates at a rate of 0.600 m/s2 for 3.50 s. What is the tension in the cable (in N) during deceleration?
(d) How high has the elevator moved above its original starting point, and what is its final velocity? (Enter the height in m and the final velocity in m/s.)
The tension in the cable supporting the elevator is 1900 N. The tension in the cable supporting the elevator during constant velocity is 15520 N. The tension in the cable supporting the elevator during deceleration is 14680 N. The elevator has moved 2.73 m above its original starting point, and its final velocity is 2.1 m/s.
(a) The acceleration is given as 1.20 m/s² and
time t = 1.75 s.
To find the tension in the cable supporting the elevator we use the formula:
Tension = mass × acceleration
Tension = 1583 × 1.2
Tension = 1899.6 N
Tension ≈ 1900 N
Therefore, the tension in the cable supporting the elevator is 1900 N.
(b) The elevator moves upward at constant velocity, so the net force acting on it is zero. Hence the tension in the cable supporting the elevator is equal to the weight of the elevator, which is given by:
Tension = mass × g
Tension = 1583 × 9.8
Tension = 15520.4 N
Tension ≈ 15520 N
Therefore, the tension in the cable supporting the elevator during constant velocity is 15520 N.
(c) During deceleration, the acceleration is negative and its magnitude is given as 0.600 m/s².
The tension in the cable supporting the elevator is given by:
Tension = mass × (g - acceleration)
Tension = 1583 × (9.8 - 0.6)
Tension = 14680.4 N
Tension ≈ 14680 N
Therefore, the tension in the cable supporting the elevator during deceleration is 14680 N.
(d) Using the formula:v = u + at
The final velocity (v) of the elevator can be calculated as:
v = u + at
v = 0 + 1.2 × 1.75
v = 2.1 m/s
To find the height the elevator has moved, we use the formula:
s = ut + 1/2 at²
The initial velocity (u) of the elevator is 0 and the time taken to reach the final velocity (v) is 1.75 s.
Therefore,s = (1/2) × 1.2 × (1.75)²
s = 2.73125 m
s ≈ 2.73 m
Thus, the elevator has moved 2.73 m above its original starting point, and its final velocity is 2.1 m/s.
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1. Choose an unusual animal sense and compare it to human senses.
Address the following in your post:
a. What kind of energy does the sense that you chose transduce? light, sound waves, infrared waves etc...
b. What part of the sense is the receptor? (What part is actually doing the transducing of energy in the environment into nervous
system impulses?)
c. Do humans have a comparable sense to this animal one? Compare the animal and human senses. In what ways are they alike? How are they different? Does every creature on our planet have this sense?
There are various animals with unusual senses that humans don't possess. However, an interesting example of an unusual animal sense is the electroreception ability found in certain species such as sharks and platypuses. Electroreception is the ability to perceive electrical fields in the environment. It is different from human senses like sight and hearing, and it is fascinating in how it works.
Addressing the given points:
a. Electroreception is the ability to sense the electrical fields that are created by living organisms or environmental sources. These animals can transduce electrical energy into nervous system impulses. Sharks, for example, use a system of jelly-filled canals and pores on their snouts called the ampullae of Lorenzini, which help them detect electric fields.
b. The receptor for electroreception is an electroreceptor organ, which is the part of the sense that actually transduces electrical energy from the environment into nervous system impulses. The organs can be found in various parts of the animal's body, such as the snout, mouth, or body surface, depending on the species.
c. Humans do not possess electroreception, so this sense is unique to animals that have evolved it. However, there are some similarities between electroreception and human senses like touch and hearing. These senses also rely on specialized receptors in the skin or ears, respectively, to transduce different types of energy (such as pressure waves or mechanical vibrations) into nervous system impulses.
In conclusion, not all creatures on our planet have this sense. Electroreception is a specialized ability that has evolved in some species to help them navigate their environment and detect prey or predators. Although humans don't have electroreception, we do have other specialized senses that help us survive and interact with the world around us.
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part 1 of 2 1. 6.40037 A 26 kg block slides down a frictionless slope which is at angle θ=28
∘
. Starting from 2. 3.95101 rest, the time to slide down is t=1.94 s. The acceleration of gravity is 9.8 m/s
2
. 3. 9.6721 4. 7.09055 5. 8.65783 6. 5.9233 7. 4.76882 What total distance s did the block slide? Answer in units of m. 8. 7.58912 Answer in units of m 9. 4.29407 10. 6.97977 1. 1.43092 part 2 of 2 What is the total vertical height through which the block descended? 2. 2.77438 Answer in units of m. 3. 4.06461 Answer in units of m 4. 2.65795 5. 3.26974 6. 2.97225 7. 2.02589 8. 4.23801 9. 2.09051 10. 3.0195
Given: A 26 kg block slides down a frictionless slope which is at angle θ=28 ∘ . Starting from rest, the time to slide down is t=1.94 s. The acceleration of gravity is 9.8 m/s2.The block slides down with uniform acceleration.
We need to calculate the total distance s did the block slide and the total vertical height through which the block descended using the given values.
1. Calculation of the distance s the block slide:
Let's use the third equation of motion,i.e. s = ut + 1/2 at²Where,u = initial velocity = 0a = acceleration = gs = ?t = 1.94 s
Putting the given values, we have:s = 0 × 1.94 + 1/2 × 9.8 × (1.94)²= 18.7717 m
Thus, the total distance s the block slide is 18.7717 m.
2. Calculation of the total vertical height:
Let's consider the right-angled triangle below: [tex]\frac{block}{height}[/tex]Thus, tan θ = opposite side / adjacent side
Hence, opposite side = adjacent side × tan θ= s × tan θ= 18.7717 × tan 28°= 10.1497 m
Thus, the total vertical height through which the block descended is 10.1497 m.
Hence, the options that answer the above two questions are:
Total distance s did the block slide = 18.7717 m.
Total vertical height through which the block descended = 10.1497 m.
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You are standing 24.1 meters away from Brown Hall. After your physics exam you want to kick a ball at the building. You kick the ball with an initial velocity of 26.7 m/s and at an angle of 33 degrees above the horizontal. Give two decimal places for your answers. What is the x-component of the initial velocity, ∼m/s What is the y-component of the initial velocity, X m/s How much time does it take for the ball to reach the building? - seconds How high up the wall, does the ball hit the building? x meters
Answer:
I apologize, it looks like my previous response was cut off. Here are the full answers to the questions:
The x-component of the initial velocity is given by:
Vx = V0 cosθ
where V0 is the initial velocity and θ is the angle above the horizontal. Substituting the given values, we get:
Vx = 26.7 cos(33°) = 22.35 m/s (to two decimal places)
Therefore, the x-component of the initial velocity is approximately 22.35 m/s.
The y-component of the initial velocity is given by:
Vy = V0 sinθ
Substituting the given values, we get:
Vy = 26.7 sin(33°) = 14.13 m/s (to two decimal places)
Therefore, the y-component of the initial velocity is approximately 14.13 m/s.
To find the time taken for the ball to reach the building, we can use the equation for the time of flight of a projectile:
t = 2Vy / g
where g is the acceleration due to gravity. Substituting the given values, we get:
t = 2(14.13) / 9.8 = 2.88 seconds (to two decimal places)
Therefore, it takes approximately 2.88 seconds for the ball to reach the building.
Tofind the height at which the ball hits the building, we can use the equation:
y = h + Vy t - 0.5 g t^2
where h is the initial height of the ball (which we can assume is zero), and y is the vertical distance traveled by the ball. Substituting the given values, we get:
y = 0 + 14.13(2.88) - 0.5(9.8)(2.88)^2 = 18.05 meters (to two decimal places)
Therefore, the ball hits the building at a height of approximately 18.05 meters above the ground.
Explanation:
Standing waves of frequency 57 Hz are produced on a string that has mass per unit length 0.0160 kg/m. With what tension must the string be stretched between two supports if adjacent nodes in the standing wave are to be 0.71 meters apart?
The tension for a desired standing wave, use the wave equation and wave velocity equation. Given the distance between adjacent nodes and frequency, the tension is approximately 105.33 Newtons.
The tension required to produce the desired standing wave, we can use the wave equation:
v = √(F/μ)
where v is the wave velocity, F is the tension in the string, and μ is the linear mass density of the string.
The wave velocity is given by the equation:
v = λf
where λ is the wavelength and f is the frequency of the wave.
In the standing wave pattern, the distance between adjacent nodes is equal to half a wavelength. So, if adjacent nodes are 0.71 meters apart, the wavelength is 2 * 0.71 = 1.42 meters.
Substituting the values into the wave velocity equation, we have:
v = λf
v = 1.42 * 57
v ≈ 81.54 m/s
Now, we can rearrange the wave equation to solve for tension:
F = μv²
Substituting the values:
F = 0.0160 * (81.54)²
F ≈ 105.33 N
Therefore, the tension required to produce the desired standing wave is approximately 105.33 Newtons.
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A bullet of mass m = 8.00 g is fired into a block of mass M = 260 g that is initially at rest at the edge of a table of height h = 1.00 m (see figure below). The bullet remains in the block, and after the impact the block lands d = 2.10 m from the bottom of the table. Determine the initial speed of the bullet.
The initial speed of the bullet is approximately 1.622 m/s.
Step 1: Apply the principle of conservation of momentum before and after the impact:
Before the impact: The momentum of the bullet is equal to the negative momentum of the block.
m * v = -(m + M) * vf
Step 2: Apply the principle of conservation of mechanical energy:
The initial potential energy of the bullet is m * g * h.
The final kinetic energy of the bullet-block system is (m + M) * vf^2 / 2.
m * g * h = (m + M) * vf^2 / 2
Step 3: Substitute the known values:
m = 8.00 g = 0.008 kg
M = 260 g = 0.260 kg
h = 1.00 m
Step 4: Solve the equations simultaneously:
From the momentum conservation equation: m * v = -(m + M) * vf
0.008 * v = -(0.008 + 0.260) * vf
0.008v = -0.268vf (equation 1)
From the energy conservation equation: m * g * h = (m + M) * vf^2 / 2
0.008 * 9.8 * 1.00 = (0.008 + 0.260) * vf^2 / 2
0.0784 = 0.268 * vf^2 (equation 2)
Step 5: Solve equation 1 for vf:
vf = -(0.008v) / 0.268 (equation 3)
Step 6: Substitute equation 3 into equation 2 and solve for v:
0.0784 = 0.268 * [-(0.008v) / 0.268]^2
0.0784 = 0.008 * v^2 / 0.268
v^2 = 0.0784 * 0.268 / 0.008
v^2 = 2.632
v = √2.632
v ≈ 1.622 m/s
Therefore, the initial speed of the bullet is approximately 1.622 m/s.
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a. Explain the meaning of the symbol on the left of the letter B in the diagram above. (1) b. State in which direction the force F acts. (2) c. Calculate the magnitude of the force F on the wire if the strength of the uniform magnetic field surrounding the current carrying wire is 420mT, the current is 13 A and 12 cm of the wire is experiencing this field. (3)
(a) The symbol on the left of the letter B in the diagram represents a uniform magnetic field.
(b) The force F acts perpendicular to both the direction of the current and the magnetic field.
(c) The magnitude of the force F on the wire can be calculated using the equation F = BIL, where B is the magnetic field strength, I is the current, and L is the length of the wire segment in the magnetic field.
(a) The symbol on the left of the letter B in the diagram represents a uniform magnetic field. A uniform magnetic field means that the magnetic field strength is constant throughout the region under consideration.
(b) According to the right-hand rule for magnetic fields, the force F on a current-carrying wire is perpendicular to both the direction of the current and the magnetic field. Therefore, the force F acts perpendicular to the plane of the diagram, either into or out of the page.
(c) The magnitude of the force F on the wire can be calculated using the equation F = BIL, where B is the magnetic field strength, I is the current flowing through the wire, and L is the length of the wire segment that is experiencing the magnetic field. Substituting the given values of B = 420 mT (or 0.420 T), I = 13 A, and L = 12 cm (or 0.12 m), we can calculate the magnitude of the force F using F = (0.420 T)(13 A)(0.12 m). Evaluating this expression gives the magnitude of the force F.
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A radioactive nucleus has a half-life of 5×10^8 years. Suppose a sample of rock (say, in an asteroid) solidified right after the solar system formed. Then approximately what fraction of the radioactive element should be left in the rock today?
Given that a radioactive nucleus has a half-life of 5 × 108 years. Let's suppose that a sample of rock (say, in an asteroid) solidified right after the solar system formed.
Then we have to calculate the fraction of the radioactive element that should be left in the rock today.
Half-life (t₁/₂) of a radioactive substance is defined as the time taken by a substance to reduce to half its initial value.
This is given by the formula,N(t) = N₀(1/2)⁽ᵗ/ᵗ₁/₂⁾ Where,N(t) = Final quantity N₀ = Initial quantity t = Time elapsed t₁/₂ = Half-life period.
We know that the half-life (t₁/₂) of the radioactive nucleus is 5 × 108 years. Hence, the fraction of the radioactive element left can be calculated as follows:After the first half-life, the quantity of the radioactive element left would be N₀/2.
After the second half-life, it would be N₀/4 and so on.
Thus, the general formula for the quantity of the radioactive element left would be,N = N₀ (1/2)n Where n is the number of half-lives elapsed.
The fraction of the radioactive element left is given as,N/N₀ = (1/2)n.
Now, we can substitute the values in the above formula.
Let's suppose that one-half-life is 5 × 108 years. Then the age of the rock would be approximately 4.6 × 109 years (age of the Solar System).
Thus, the number of half-lives elapsed would be given by,n = (time elapsed)/(half-life)n = (4.6 × 109)/(5 × 108) = 9.2.
After 9.2 half-lives, the fraction of the radioactive element left would be,N/N₀ = (1/2)⁹.²≈ 0.00077 ≈ 7.7 × 10⁻⁴.
Thus, approximately 0.077% (7.7 × 10⁻⁴) of the radioactive element should be left in the rock today.
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A layer of ice having parallel sides floats on water. If light is incident on the upper surface of the ice at an angle of incidence of 26.6° , what is the angle of refraction in the water? Noed Help? Restit block). A fraction of the light is reflected and the rest refracted. What is the angle (in degrees) between the refiected and refracted rays?
The angle of refraction in the water is approximately 20.83°, and the angle between the reflected and refracted rays is approximately 32.47°.
To determine the angle of refraction in the water and the angle between the reflected and refracted rays, we can use Snell's law, which relates the angles of incidence and refraction at an interface between two mediums. The law is stated as:
n₁ × sin(θ₁) = n₂ × sin(θ₂)
Where:
n₁ is the refractive index of the initial medium (in this case, air)
θ₁ is the angle of incidence
n₂ is the refractive index of the second medium (in this case, water)
θ₂ is the angle of refraction
In this case, since the incident medium is air and the second medium is water, we can assume the refractive index of air to be approximately 1 and the refractive index of water to be around 1.33.
Given that the angle of incidence (θ₁) is 26.6°, we can calculate the angle of refraction (θ₂) as follows:
1 × sin(26.6°) = 1.33 × sin(θ₂)
sin(θ₂) = (1 × sin(26.6°)) / 1.33
θ₂ = arcsin((1 × sin(26.6°)) / 1.33)
Using a calculator, we can find that θ₂ is approximately 20.83°.
Now, to calculate the angle between the reflected and refracted rays, we can use the fact that the angle of incidence is equal to the angle of reflection. Therefore, the angle between the reflected and refracted rays will be:
Angle between reflected and refracted rays = 2 × θ₁ - θ₂
Angle between reflected and refracted rays = 2 × 26.6° - 20.83°
Using a calculator, we can find that the angle between the reflected and refracted rays is approximately 32.47°.
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A crate of mass m1 = 14.80 kg is pulled by a massless rope up a 36.9° ramp. The rope passes over an ideal pulley and is attached to a hanging crate of mass m2 = 16.30 kg. The crates move 1.39 m, starting from rest.
Find the work done by gravity on the sliding crate. J
The work done by gravity on the sliding crate is approximately 147.55 Joules.
To find the work done by gravity on the sliding crate, we need to calculate the change in gravitational potential energy.
The gravitational potential energy is given by the formula:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the vertical height.
In this case, the sliding crate moves up the ramp, so we need to consider the change in height along the incline.
The change in height, Δh, can be calculated using trigonometry:
Δh = d * sin(θ)
where d is the distance the crate moves along the ramp and θ is the angle of the ramp.
Mass of sliding crate, m1 = 14.80 kg
Mass of hanging crate, m2 = 16.30 kg
Angle of the ramp, θ = 36.9°
Distance moved along the ramp, d = 1.39 m
Acceleration due to gravity, g = 9.8[tex]m/s^2[/tex]
First, calculate the change in height:
Δh = 1.39 m * sin(36.9°)
Next, calculate the work done by gravity:
Work = ΔPE = m1 * g * Δh
Substituting the values, we have:
Work = 14.80 kg * 9.8 [tex]m/s^2[/tex] * Δh
Calculate Δh and substitute the value:
Work = 14.80 kg * 9.8[tex]m/s^2[/tex] * (1.39 m * sin(36.9°))
Finally, calculate the value:
Work ≈ 147.55 J
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1 light-second in kilometers express your answer using three significant figures.
One light-second in kilometers can be expressed using three significant figures as 299,792 kilometers.
This value represents the distance that light travels in one second in a vacuum. In other words, light travels at a constant speed of 299,792 km/s in a vacuum. Therefore, one light second is equivalent to this distance. This conversion factor is useful in various fields of science, such as astronomy and telecommunications.
To obtain this answer, we can use the exact speed of light, which is 299,792,458 meters per second. Since we need to convert it to kilometers, we divide this value by 1,000, which gives us 299,792.458 kilometers per second.
Rounding off this value to three significant figures, we get 299,792 kilometers per second. Finally, to get the distance that light travels in one second, we multiply this value by one, which gives us 299,792 kilometers (rounded to three significant figures).
Therefore, 1 light-second is equal to 299,792 kilometers (rounded to three significant figures).
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A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of 59.7 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road. v=
Let v be the maximum speed that the motorcycle can have while moving over the crest without losing contact with the road.
Since the hill's crest is a circular arc with a radius of 59.7 m,
its weight W can be resolved into two components: a radial force W cos θ that is perpendicular to the road and a tangential force W sin θ that is parallel to the road.Let's now take a look at the forces acting on the motorcycle. The forces that act on the motorcycle are the gravitational force W, the centripetal force F, and the force of friction f.
As a result, the following equation can be used to find the maximum speed that the motorcycle can have while moving over the crest without losing contact with the road:
`Ff = mv²/r`where `m` is the mass of the motorcycle and `r` is the radius of the circular arc of the hill.
We can calculate the radial component of the weight as
`W cos θ = mg cos θ`, where `m` is the mass of the motorcycle and `g` is the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting `W cos θ` and `W sin θ` into the equation for `Ff`, we have:
`f = µW cos θ` and `F = W sin θ`
Substituting these equations into the equation for `Ff`, we have:
`µmg cos θ = mv²/r - mg sin θ`
Simplifying this expression yields:
`v² = rg(µ cos θ - sin θ)`
Substituting the given values, we have:
`v² = (59.7 m)(9.8 m/s²)(0.9) = 522.7 m²/s²`
Therefore, the maximum speed that the cycle can have while moving over the crest without losing contact with the road is:
`v = sqrt(522.7 m²/s²) = 22.85 m/s`
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Observing that the ball rolls down the inclined plane, determine what the acceleration of the ball is as it rolls (assuming no friction) down the ramp. Note, you may be tempted to answer, "the acceleration of the ball is caused by the acceleration due to gravity which is 9.8 m/s?, however notice the ball does not fall vertically downward. Using the inclined plane as a right triangle, use trig to determine what the acceleration of the ball is. You will need to know the angle of inclination of the plane, which you can find using the images above.
To determine the acceleration of a ball as it rolls down an inclined plane (assuming no friction), we need to use trigonometry. We need to find the component of the force due to gravity that pulls the ball down the ramp. The acceleration of the ball is equal to this component divided by the mass of the ball.The angle of inclination of the plane is given as 30°.From the image, we see that the force due to gravity can be split into two components:
one parallel to the ramp (Fp) and one perpendicular to the ramp (Fn).The force parallel to the ramp (Fp) is given by Fp = mgsinθ, where m is the mass of the ball, g is the acceleration due to gravity, and θ is the angle of inclination of the plane.The force perpendicular to the ramp (Fn) is given by Fn = mgcosθ, where m is the mass of the ball, g is the acceleration due to gravity, and θ is the angle of inclination of the plane.The acceleration of the ball down the ramp is given by a = Fp/m. We can substitute Fp into this equation, giving us a = mgsinθ/m = gsinθ.Using the given angle of inclination of the plane (θ = 30°) and the acceleration due to gravity (g = 9.8 m/s²), we can calculate the acceleration of the ball as it rolls down the ramp:
a = gsinθ = 9.8 m/s² × sin(30°) ≈ 4.9 m/s²Therefore, the acceleration of the ball as it rolls down the inclined plane is approximately 4.9 m/s².About GravityGravity is a natural phenomenon whereby everything that has mass or energy in the universe—including planets, stars, galaxies, and even light—attracts one another. Gravity is useful for holding objects on the surface of the earth. If there is no gravitational force, objects will scatter and collide with each other. Objects on earth can also be thrown into space. The force of gravity keeps the atmosphere on the earth's surface.
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The emf of the battery in the circuit below is 30 V and the internal resistance of the battery is zero. R _1 =2 Ohms and
R _2 =1 Ohms: Find the total resistance of the network. Find the current flowing through the battery in the circuit.
The total resistance of the network is 3 Ohms. The current flowing through the battery in the circuit is 10 Amperes.
To find the total resistance of the network, we can use the formula for resistors in series:
R_total = R_1 + R_2
R_1 = 2 Ohms
R_2 = 1 Ohm
Substituting the given values into the formula:
R_total = 2 Ohms + 1 Ohm
R_total = 3 Ohms
Therefore, the total resistance of the network is 3 Ohms.
To find the current flowing through the battery in the circuit, we can use Ohm's Law:
I = V / R
I is the current
V is the voltage (emf) of the battery
R is the total resistance of the network
V = 30 V
R = 3 Ohms
Substituting the given values into the formula:
I = 30 V / 3 Ohms
I = 10 Amperes
Therefore, the current flowing through the battery in the circuit is 10 Amperes.
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An at-track gider is attached to a spring The glider is pulled to the right and released from rest at f=0 then states with a period of 1.8s and a maximum speed of 44 cm/s 96 5 R E D C LL 1 T V 6 G & Y B N H Part A What is the amplitude of the oscilation? Express your answer in centimeters Subm Part B What is the per pot247 Express your aris cantimeters Ale A 0 U 8 00 N VAC J ( 9 O M O L K
Part A: To determine the amplitude of the oscillation, we can use the relationship between the maximum speed and the amplitude of simple harmonic motion. The maximum speed of the glider is given as 44 cm/s. The maximum speed occurs at the amplitude of the oscillation. Therefore, the amplitude of the oscillation is 44 cm.
Part B: The period of the oscillation is given as 1.8 s. The period (T) is the time taken for one complete cycle of the oscillation. The frequency (f) is the reciprocal of the period, so we have f = 1/T. Substituting the given value, we have f = 1/1.8 s ≈ 0.556 Hz.
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Determine the one or more conditions required for the linear momentum in a system to have reached steady state : A. There are no external forces but mass can be transported into or out of the system B. the system has constant acceleration and constant mass C. No mass is transported into or out of the system but external forces can be applied D. the system has constant velocity and constant mass The rate form of the conservation of linear momentum reduces to Newton's second law under what condition(s): Select one or more of the answers below A. Min = 0 B. Mout = 0 oc. Fnet = 0 D.ag=0 (G refers to the center of mass) E. m sys=0
C. No mass is transported into or out of the system but external forces can be applied
In steady state, the system reaches a balance where the mass within the system remains constant, but external forces can still act on the system.
The rate form of the conservation of linear momentum reduces to Newton's second law under the condition(s):
D. Fnet = 0 (Net external force acting on the system is zero)
When the net external force acting on the system is zero, the rate form of the conservation of linear momentum reduces to Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration.
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If a 220 V step down transformer is used for lighting eight 12 V, 20 W lamps, find the efficiency of the transformer when a current of 1 A exists in the primary coil.
If a 220 V step down transformer is used for lighting eight 12 V, 20 W lamps , the efficiency of the transformer is 72.73%.
A transformer can be described as a static electrical device that transfers electrical energy from one circuit to another through electromagnetic induction. The primary and secondary coils are the two main components. The efficiency of the transformer is the ratio of the output power to the input power.
The given data are: Primary voltage, V1 = 220 V
Primary current, I1= 1 A
Secondary voltage, V2 = 12 V
Power of each lamp, P = 20 W
Number of lamps, n = 8
The primary power is given by P1 = V1I1 = 220 × 1 = 220 W .
The secondary current is calculated as,
I2 = P/nV = 20/(12 × 8) = 0.2083 A.
The secondary power is given by P2 = nPI2 = 8 × 20 = 160 W.
Therefore, the efficiency of the transformer is given by η = P2/P1× 100= 160/220 × 100 = 72.73%.
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a
0.0kg cylinder has a radius 0.2m and a torque of 0.0 N.m applied to
the shaft. determine the rotational speed of the cylinder after 5
s, starting from rest
The initial angular velocity (ω1) of the cylinder is zero.
The angular acceleration (α) is unknown.
The torque (τ) acting on the cylinder is 0 N.m.
The mass (m) of the cylinder is 0.0 kg.
The radius (r) of the cylinder is 0.2 m
. The moment of inertia (I) of a solid cylinder is (1/2)mr2.
Thus: I = (1/2)(0.0 kg)(0.2 m)2 = 0 J.s2.
To determine the final angular velocity (ω2) of the cylinder after 5 s, we use the equation:
ω2 = ω1 + αtω2 = 0 + α(5)ω2 = 5αTo determine the angular acceleration (α), we use the equation:
τ = Iα0 = (1/2)(0.0 kg)(0.2 m)2αα = 0 N.m / (1/2)(0.0 kg)(0.2 m)2α = 0 N.m / 0 J.s2α = undefined
Substituting the value of α into the equation for ω2:
ω2 = 5αω2 = 5(undefined)ω2 = undefined
The final angular velocity of the cylinder cannot be determined, as the angular acceleration is undefined. Therefore, the cylinder will not rotate.
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A certain electromagnetic wave source operating at 10 W output power emits EM waves at the frequency of 4.59 x 1014 Hz. How many photons are emitted by this source over a period of 1 minute?
The photon energy formula is given as `E=hf`, where `E` is energy, `h` is Planck's constant, and `f` is frequency. Hence, the number of photons `n` emitted by the electromagnetic wave source over a period of 1 minute is given by:`
n = (Power x Time) / Energy of 1 photon`In this question, the output power of the electromagnetic wave source is 10 W and the frequency of the EM waves emitted by the source is 4.59 x 1014 Hz.To calculate the energy of 1 photon, we use the photon energy formula:`E = hf = (6.626 x 10^-34 J s) x (4.59 x 10^14 Hz) = 3.042 x 10^-19 J`Therefore, the number of photons emitted by the source over a period of 1 minute is:`n = (Power x Time) / Energy of 1 photon``n = (10 W x 60 s) / (3.042 x 10^-19 J)`n = 1.97 x 10^22 photons (approx.)Therefore, the electromagnetic wave source emits approximately `1.97 x 10^22` photons over a period of 1 minute.
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The electric field of an electromagnetic wave traveling in vacuum is described by the following wave function: E=(5.00 V/m)sin[kx−(6.00×109 s−1)t]^ In this equation, k is the wave number in rad/m,x is in m, and t is in s. Assume that ^,^, and k^ are the unit vectors along x-axis, y− axis, and z− axis, respectively. Find the following quantities: (a) amplitude (b) frequency (c) wavelength (d) the direction of the travel of the EM wave (e) the equation of the magnetic field with correct unit vector
Summary:
(a) The amplitude of the electric field is 5.00 V/m.
(b) The frequency of the electromagnetic wave is 6.00 × 10^9 s^(-1) or 6.00 GHz.
(c) The wavelength of the electromagnetic wave is approximately 5.00 × 10^(-4) m or 0.50 mm.
(d) The direction of travel of the electromagnetic wave is along the positive x-axis.
(e) The equation of the magnetic field can be determined by relating it to the electric field and the wave's speed.
(a) The amplitude of the electric field, E_0, is given as 5.00 V/m in the wave function: E = (5.00 V/m)sin[kx - (6.00 × 10^9 s^(-1))t].
(b) The frequency, f, of the electromagnetic wave can be determined from the angular frequency, ω, using the relationship ω = 2πf. In this case, ω = 6.00 × 10^9 s^(-1), so solving for f gives f = ω / (2π) = (6.00 × 10^9 s^(-1)) / (2π) ≈ 9.55 × 10^8 Hz or 955 MHz.
(c) The wavelength, λ, of the wave can be determined from the wave number, k, using the relationship k = 2π / λ. Rearranging the equation, we find λ = 2π / k. In this case, k is not provided explicitly, so we cannot determine the wavelength accurately without knowing its value.
(d) The direction of travel of the electromagnetic wave is determined by the sign of the coefficient of the x-term in the wave function. In this case, the coefficient is positive, indicating that the wave is propagating along the positive x-axis.
(e) The equation of the magnetic field, B, can be determined using the relationship between the electric field, E, and the magnetic field, B, in an electromagnetic wave: B = (E / c) × n, where c is the speed of light in vacuum and n is the unit vector in the direction of propagation. Since the wave is traveling in vacuum, c = 3.00 × 10^8 m/s. Therefore, the equation of the magnetic field is B = (5.00 V/m) / (3.00 × 10^8 m/s) × k^, where k^ is the unit vector along the z-axis.
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an electrically charged object can be used to attract:
An electrically charged object can be used to attract any object with an opposite charge.
This is due to the fundamental principle that opposites attract and repel in physics.
Electric charge is a fundamental property of matter that gives rise to electromagnetic interactions. An electric charge, whether positive or negative, produces an electric field that surrounds it. This field exerts a force on any other charge in its vicinity that is either attracted to or repelled from it. Electric charge is a fundamental property of matter that produces a variety of electric phenomena. When the charge is concentrated in a localized region of space, the object is electrically charged. When there is a net accumulation of charge in an object, it becomes electrically charged. An electrically charged object produces an electric field in its vicinity, which exerts a force on other charged objects. An electrically charged object can be used to attract objects with an opposite charge or repel objects with the same charge.
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Star A has a magnitude of 6 and Star B has a magnitude of 15 . How much brighter is Star A than Star B? a. 1.5 b. 3815 c. 2.5 d. 2 e. 97.7 f. 0.0102 g. 6.25 h. 0.00164 i. 0.0002621 j. 5 k. 1526 I. 610 m. 0.0006554 n. 3.33 o. 0.16
The correct answer is Option f. Star A is 512.45 times brighter than Star B, or in other words, Star A is 0.0102 times as bright as Star B.
The magnitude of a star refers to its brightness as seen from Earth.
The magnitude scale is such that a difference of 1 magnitude unit is equal to a brightness difference of 2.512.
If one star has a magnitude of 6, and the other has a magnitude of 15, the difference in magnitude between them is 9 (15 - 6 = 9).
The brightness difference can be calculated using the magnitude difference between the two stars, using the following formula: Brightness difference = [tex]2.512^{(magnitude difference)}[/tex]
In this case, the magnitude difference between the two stars is 9.
So, the brightness difference can be calculated as:
[tex]Brightness difference = 2.512^9 = 512.45[/tex]
Therefore, Star A is 512.45 times brighter than Star B, or in other words, Star A is 0.0102 times as bright as Star B.
Hence, the correct answer is f. 0.0102.
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A Trumpeter is playing a note with a frequency of 565 Hz while sitting on a vehicle driving towards a large building. If the conductor, standing on the same vehicle, hears a beat frequency of 7 Hz made from the sound coming from the trumpeter and the Doppler Shifted note rebounding off the building, how fast is the vehicle moving?
The vehicle is moving at a speed of approximately 24.85 m/s.
When a source of sound, in this case, the Trumpeter, and an observer, in this case, the conductor, are in relative motion, the Doppler effect comes into play. The beat frequency heard by the conductor is the difference between the frequency emitted by the Trumpeter and the Doppler-shifted frequency of the sound reflected off the building. The beat frequency can be calculated by subtracting the Doppler-shifted frequency from the emitted frequency.
In this scenario, the beat frequency is given as 7 Hz, and the emitted frequency is 565 Hz. By solving the equation for the Doppler effect, we can determine the Doppler-shifted frequency. Since the conductor hears the beat frequency made up of the emitted frequency and the Doppler-shifted frequency, the difference between the two frequencies is equal to the beat frequency.
With the known values, we can rearrange the equation to find the speed of the vehicle. By substituting the given values into the equation, we can calculate the velocity of the vehicle.
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An electron with an initial speed of 4.80x105 m/s is brought to rest by an electric field. Did the electron move into a region of higher potential or lower potential? Lower potential Higher Potential Same potential 2 points Saved QUESTION 2 Electric field Part B What was magnitude (absolure value) of the potential difference in volts that stopped the electron (Do not enter any units)
An electron with an initial speed of 4.80x10^5 m/s is brought to rest by an electric field. The electron moved into a region of higher potential because an electric field moves charged particles from higher potential to lower potential. Since electrons have negative charges, the direction of the electric field is opposite to the direction of force on an electron.
To determine the magnitude of the potential difference in volts that stopped the electron, we can use the formula for potential difference: Potential Difference = Kinetic Energy / Charge.
The kinetic energy of the electron is given by the formula: Kinetic Energy = (1/2)mv², where m is the mass of the electron and v is its initial velocity.
The charge of an electron is -1.60 × 10^-19 C.
Substituting the values into the potential difference formula, we get: Potential Difference = [(1/2)(9.11 × 10^-31 kg)(4.80 × 10^5 m/s)²]/(1.60 × 10^-19 C) = 1.16 × 10^3 V.
Therefore, the magnitude of the potential difference in volts that stopped the electron is 1.16 × 10^3 V.
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A disk with a mass of M-10kg is supported by a frictionless axle and positioned in a vertical plane. A mass of m=120g is tied to a string and wrapped around a small groove at the edge of the disk. Determine the tension T experienced by the string in [N] after the mass is released from res. The moment of inertia is I=1/2 mr^2
To determine the tension experienced by the string, we need to consider the forces acting on the system.
When the mass m is released, it will accelerate downwards due to the force of gravity. This downward acceleration will cause a torque on the disk, which will result in angular acceleration.
The tension in the string will provide the torque necessary to accelerate the disk. The torque due to the tension can be calculated as the product of the tension T and the radius of the disk r.
The gravitational force acting on the mass m will also contribute to the torque. The weight of the mass m can be calculated as mg, where g is the acceleration due to gravity.
In rotational equilibrium, the torque due to the tension and the torque due to the weight of the mass m must balance. Therefore, we can write:
Tension × radius = Weight of mass m × radius
Solving for the tension T, we have:
T = (Weight of mass m) × (radius / radius)
Substituting the given values and performing the calculations will yield the tension T experienced by the string in newtons.
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If astronauts could travel at v = 0.956c, we on Earth would say it takes (4.20/0.956) = 4.39 years to reach Alpha Centauri, 4.20 light-years away. The astronauts disagree. (a) How much time passes on the astronauts' clocks? years (b) What is the distance to Alpha Centauri as measured by the astronauts? light-years
(a) 2.52 years pass on the astronauts' clocks during their journey to Alpha Centauri.
(b) The distance to Alpha Centauri remains 4.20 light-years as measured by the astronauts.
When objects move at speeds close to the speed of light (c), time dilation occurs due to the theory of special relativity. According to this theory, as an object's velocity approaches the speed of light, time slows down for that object relative to an observer at rest. In this case, the astronauts are traveling at a velocity of v = 0.956c, which is 95.6% of the speed of light.
(a) Due to time dilation, less time passes on the astronauts' clocks compared to an observer on Earth. To calculate the time experienced by the astronauts, we can use the time dilation formula:
Δt' = Δt / √(1 - (v²/c²))
Here, Δt represents the time measured by an observer on Earth, Δt' represents the time experienced by the astronauts, v is the velocity of the astronauts, and c is the speed of light.
Substituting the given values, we have:
Δt' = 4.20 years / √(1 - (0.956²))
Calculating this equation gives us:
Δt' = 2.52 years
Therefore, only 2.52 years pass on the astronauts' clocks during their journey to Alpha Centauri.
(b) The distance to Alpha Centauri remains the same, regardless of the astronauts' velocity. From the perspective of the astronauts, the distance is still 4.20 light-years. Length contraction is another consequence of special relativity, which implies that the length of objects moving at high speeds appears shorter when observed from a different frame of reference.
However, this contraction does not affect the actual distance between objects.
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Physical units in mechanics are usually some combination of the dimensions time T, mass M, and length L. Consider the physical quantities m,r,v,a, and t with dimensions [m]=M,[r]=L,[v]=LT−1 ,[a]=LT ^−2 , and [t]=T. Enter the dimensional expression of the quantity on the right-hand side of each equation. Your answers may contain only M, L, T, and exponents. Assume that each of the following equations is dimensionally consistent. L 0 =mvr [L1 W=mar k=− rma
The dimensional expressions for the quantities on the right-hand side of the given equations are ML²T⁰, ML²T⁻¹, and MLT⁻², corresponding to different physical quantities involved in the equations.
Physical quantities are m, r, v, a, and t with dimensions [m] = M, [r] = L, [v] = LT⁻¹, [a] = LT⁻², and [t] = T. The dimensional expression of the quantity on the right-hand side of each equation is given below:
L0 = mvr
where [L0] = L1[L] = [M]a[L]b[T]c = MaLbTc
By equating the dimensions on both sides, we get
LHS = RHS.
LHS = L0 = L¹
RHS
mvr = [M][L][LT⁻¹] = MaL²T⁻¹
Comparing the exponents of M, L, and T on both sides, we get
M : 1 = aL : 2 = bT : -1 + 1 = c⇒ a = 1, b = 2, and c = 0.
So, the dimensional expression of the quantity on the right-hand side of L0 = mvr is MaL²T⁰ = ML²T⁰W = mar
where [W] = [F][d] = MLT⁻²LT = ML²T⁻¹
By equating the dimensions on both sides, we get
LHS = RHS.
LHS = W = ML²T⁻¹
RHS
mar = [M][LT⁻²][L] = ML²T⁻¹
Comparing the exponents of M, L, and T on both sides, we get
M : 1 = 1
L : 2 = 1
T : -1 - 2 = -3⇒ the dimensional expression of the quantity on the right-hand side of W = mar is ML²T⁻¹.
K = -rma
By equating the dimensions on both sides, we get
LHS = RHS.
LHS = K = [M][L²][T⁻²]
RHS
-rma = -[L][M][T⁻²] = MLT⁻²
Comparing the exponents of M, L, and T on both sides, we get
M : 1 = 1
L : 2 = -1
T : -2 = -2⇒ the dimensional expression of the quantity on the right-hand side of K = -rma is MLT⁻².
Hence, the dimensional expression of the quantity on the right-hand side of each equation is
ML²T⁰, ML²T⁻¹, and MLT⁻².
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Each of the following statements about the electric field in a conductor at equilibrium was written by a different student. Select all hose that are physically correct. At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, so the net electric field inside the conductor is equal to the electric field due to charges in the surroundings. At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, and the electric field due to charges in the surroundings cannot penetrate the conductor, so the net electric field inside the conductor must be zero. At equilibrium the net electric field inside a conductor must be zero, because if it were not zero, there would be charge flow because the drift speed of the mobile charges is proportional the the net electric field. At equilibrium the net electric field inside a conductor must be zero, because the average drift speed of the mobile charges is
v
ˉ
=uE
net
, and the only way for
v
ˉ
to be zero is if E
net
=0. At equilibrium the electric field inside a conductor at equilibrium is zero because electric fields due to charges in the surroundings cannot penetrate the material of the conductor. At equilibrium the net electric field inside a conductor is zero because the conductor polarizes until the electric field inside the conductor due to charges at the surface is equal and opposite to the electric field due to charges in the surroundings.
At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, so the net electric field inside the conductor is equal to the electric field due to charges in the surroundings.
At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, and the electric field due to charges in the surroundings cannot penetrate the conductor, so the net electric field inside the conductor must be zero.
At equilibrium the net electric field inside a conductor must be zero, because the average drift speed of the mobile charges is v ˉ =uE net , and the only way for v ˉ to be zero is if E net =0. At equilibrium the net electric field inside a conductor is zero because the conductor polarizes until the electric field inside the conductor due to charges at the surface is equal and opposite to the electric field due to charges in the surroundings.
When an electric field is applied to a conductor, the free charges inside the conductor experience an electric force. The charges move and keep moving until the charge redistribution due to the motion of charges results in the elimination of the electric field inside the conductor.At this point, the redistribution of charges inside the conductor stops, and the conductor is said to have reached its electrostatic equilibrium.
During this equilibrium, there is no further movement of charges. Therefore, no current flows through the conductor.Therefore, only the following four statements are correct:At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, so the net electric field inside the conductor is equal to the electric field due to charges in the surroundings.
At equilibrium the electric field inside a polarized conductor due to the charges accumulated at the surface of the conductor is zero, and the electric field due to charges in the surroundings cannot penetrate the conductor, so the net electric field inside the conductor must be zero.
At equilibrium the net electric field inside a conductor must be zero, because the average drift speed of the mobile charges is v ˉ =uE net , and the only way for v ˉ to be zero is if E net =0.
At equilibrium the net electric field inside a conductor is zero because the conductor polarizes until the electric field inside the conductor due to charges at the surface is equal and opposite to the electric field due to charges in the surroundings.
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