Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :




t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

Here, a = -g


h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
The kinetic energy K given to the helium nucleus is equal to its potential energy, which is

where q=2e is the charge of the helium nucleus, and

is the potential difference applied to it.
Since we know the kinetic energy, we have

and from this we can find the potential difference:
Answer:
T = 60 s
Explanation:
There are 6 poles on the track which are equally spaced
so the angular separation between the poles is given as


so the angular speed of the train is given as


now we have time period of the train given as



Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= <u><em>- 4 m/s</em></u>
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV