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34kurt
3 years ago
10

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

axis of the loop. Somebody carefully grabs the edges of the loop and begins pulling them apart such that the area of the loop increases at a rate of 20 cm2/s. What is the magnitude of the induced EMF in the loop?
Physics
1 answer:
Fofino [41]3 years ago
7 0

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

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3 years ago
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with m_{Pa} and its initial velocity with u_{Pa}

Let represent the mass of Ricardo with m_{Ri} and its initial velocity with u_{Ri}

At rest ;

their velocities will be zero, i.e

u_{Pa} = u_{Ri} = 0

The initial momentum for this process can be represented as :

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} = 0

after push off from each other then their final velocity will be v_{Pa} and v_{Ri}

The we can say their final momentum is:

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri} = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} =  m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

Since the initial velocities are stating at rest then ; u = 0

m_{Pa}(0) + m_{Pa}(0) = m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}  = 0

m_{Pa}v_{Pa} = - m_{Ri}v_{Ri}

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So m_{Ri} > m_{Pa} ;

Then \mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

m_{Pa}v_{Pa} =  m_{Ri}v_{Ri}

The ratio is

\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1

v_{Pa} >v_{Ri}

Therefore, Paula will have a greater speed than Ricardo after the push-off.

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