KE = 1/2 mv^2
in this case, the initial kinetic energy which is converted to heat is
KE = 1/2 1400 (12)^2
KE = 100,800 J
Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :


E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Answer:
E. Student 1 is correct, because as θ is increased, h is the same.
Explanation:
Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.
Mathematically:

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.
And for work we have:

where:
displacement of the object
angle between the force and displacement vectors
Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.
- In case of free-fall the angle between the force is and the displacement is zero.
- In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.
So, the work done by the gravitational force is same in the two cases.
The answer is "Three".
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