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nikdorinn [45]
3 years ago
10

An elastic conducting material is stretched into a circular loop of 10.7 cm radius. It is placed with its plane perpendicular to

a uniform 0.803 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 81.0 cm/s. What emf is induced in volts in the loop at that instant?
Physics
1 answer:
valina [46]3 years ago
3 0

Answer:

0.44 V

Explanation:

Parameters given:

Radius of loop, r = 10.7 cm = 0.107 m

Magnetic field strength, B = 0.803 T

Rate of shrinkage of Radius, dr/dt = 81 cm/s = 0.81m/s

EMF induced is given in terms of magnetic Flux, Φ, as:

EMF = dΦ/dt

Magnetic Flux, Φ, is given as:

Φ = B * A (where A is area of loop)

Therefore, EMF is:

EMF = d(B*A)/dt

The area of the loop is given as A = πr²

EMF = d(Bπr²) / dt = Bπ*d(r²)/dt

=> EMF = Bπ*2r*(dr/dt)

EMF = 0.803 * π * 2 * 0.107 * 0.81

EMF = 0.44 V

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The force of friction.

Explanation:

Gravity keeps the car on the ground.

Motion Allows the car to move.

The force of speed doesnt make sense.

Friction would cause the car to stop moving.

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A radioactive isotope of the element potassium decays to produce argon. If the ratio of argon to potassium is found to be 31:1,
iris [78.8K]

Answer:

Explanation:

Argon to potassium ratio after 1 half life = 1:1

After 2 half lives = 75/25= 3:1

After 3 half lives = 87.5/12.5= 7:1

After 4 half lives = 93.75/6.25 = 15:1

After 5 half lives = 96.875/3.125 = 31/1

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3 years ago
Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i
hammer [34]

Answer:

a_M=1.92a_A

Explanation:

\omega_M=\omega_A = Angular speed

r_M = Distance of Mary = 11.5 ft

r_A = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A

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8 0
3 years ago
A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq
Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>
  • The following expression can be used to relate the fundamental frequency to the driving frequency;

                                        f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

  • The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

                                          f(4) = 4× f (1)

                                          480 = 4× f(1)

                                         f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

  • To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
  • For, n = 5,

                      f(n) = n 120Hz,

                      f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

#SPJ4

8 0
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Answer:

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v = 2.5 m/s

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2 years ago
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