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Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Answers

Answer 1
Answer:

Answer:

0.00124 V

Explanation:

Parameters given:

Initial circumference = 162 cm

Rate of decrease of circumference = 14 cm/s

Magnetic field, B = 0.5 T

Time, t = 8 secs

The magnitude of the EMF induced in the loop is given as:

V = (-NBA) / t

Where N = number of turns = 1

B = magnetic field

A = area of loop

t = time taken

First, we need to find the area of the loop.

To do this, we will find the radius after the loop circumference has decreased for 8 secs.

The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:

14 * 8 = 112 cm

The new circumference is:

162 - 112 = 50 cm = 0.5 m

To get radius:

C = 2 * pi * r

r = C / (2 * pi)

r = 0.5 / (2 * 3.142)

r = 0.0796 m

The area is:

A = pi * r²

A = 3.142 * 0.0796²

A = 0.0199 m²

Therefore, the EMF induced is:

V = (-1 * 0.5 * 0.0199) / 8

V = -0.00124V

This is the EMF induced in the coil.

The magnitude is |-0.00124| V = 0.00124 V.

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Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer:

4U_0

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P=U_0

We know that potential at point P=V=(U)/(q)

U=Vq

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially ,V=(U_0)/(q_0)

New charge, q=4q_0

Then, work done,U=(U_0)/(q_0)* (4q_0)=4U_0

Hence, the amount of  work needed to bring a new charge 4q0 from infinity to point P=4U_0

Answer:

4U_(0)

Explanation:

V = u / q,

Work = P = V

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What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? A. Use a lens to focus the power into a smaller area. B. Increase the power output of the lamp. C. Either A or B.

Answers

Answer:

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d 5.2 m/s

Answers

Answer:

The pin's final velocity is 5m/s

Explanation:

Step one:

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Explain the difference between kingdom and a species

Answers

The Kingdom is the highest rank in classification. It is subdivided into 5 major kingdoms while as species is the basic unit of classification. A species can be defined as a group of interbreeding organisms that are capable of producing fertile offspring.

A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

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where:

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m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_(1)*v_(1))+F*t=(m_(1)*v_(2))

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

(10*1)+F*5=10*4\n10+5*F=40\n5*F=40-10\n5*F=30\nF=6[N]
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