Answer:
I would say the answer is A... but I'm not so sure ....
Answer:
Sorry I don't know the answer .Hope other help you.sorryyyyyyyy
Answer:
a. 20m/s
b.50N
c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.
Explanation:
a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

Hence, the velocity of the ball after the kick is 20m/s
b.The force felt by the turkey:
#Applying Newton's 3rd Law of motion, opposite and equal reaction:
-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.
c. Using the law of momentum conservation:
-Due to ther external forces exerted on the turkey, it remains stationery.
-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.
-Momentum is not conserved due to these external forces.
Answer:
Part a)

Part B)

Part C)

Explanation:
Part a)
Magnetic field due to a long ideal solenoid is given by

n = number of turns per unit length



now we know that magnetic field due to solenoid is


Now magnetic flux due to this magnetic field is given by




Part B)
Now for mutual inductance we know that




now we have


Part C)
As we know that induced EMF is given as



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