Answer:
I=1.48 A
Explanation:
Given that
B=3.1 x 10⁻5 T
b= 4.2 cm
l= 9.5 cm
The relationship for magnetic field and current given as

Where

By putting the values


D=26.03 m⁻¹



I=1.48 A
Answer: 247.67 V
Explanation:
Given
Potential At A 
Potential at 
when particle starts from A it reaches with velocity
at Point while when it starts from C it reaches at point B with velocity 
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B

Change in Kinetic Energy of particle moving under the Potential From C to B

Divide 1 and 2 we get

on solving we get


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Answer:
47.4 m
Explanation:
When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.
In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

where
s is the vertical displacement
u = 0 is the initial velocity
t = 3.11 s is the time
is the acceleration of gravity (taking downward as positive direction)
Solving the formula, we find
