Explanation:
It is given that,
Mass of the ball, m = 0.06 kg
Initial speed of the ball, u = 50.4 m/s
Final speed of the ball, v = -37 m/s (As it returns)
(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :
J = -5.24 kg-m/s
(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.
W = -35.1348 Joules
Hence, this is the required solution.
Answer:
acceleration
acceleration is the rate at which velocity change
i think
One quater as your moving away more!
Answer:
The coil radius of other generator is 5.15 cm
Explanation:
Consider the equation for induced emf in a generator coil:
EMF = NBAω Sin(ωt)
where,
N = No. of turns in coil
B = magnetic field
A = Cross-sectional area of coil = π r²
ω = angular velocity
t = time
It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.
Therefore, EMF for both coils or generators will be:
EMF₁ = NBπr₁²ω Sin(ωt)
EMF₂ = NBπr₂²ω Sin(ωt)
dividing both the equations:
EMF₁/EMF₂ = (r₁/r₂)²
r₂ = r₁ √(EMF₂/EMF₁)
where,
EMF₁ = 1.8 V
EMF₂ = 3.9 V
r₁ = 3.5 cm
r₂ = ?
Therefore,
r₂ = (3.5 cm)√(3.9 V/1.8 V)
<u>r₂ = 5.15 cm</u>