A phase of moon with more than half sunlit portion visible on the right
Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J
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They both Move energy through a Field without Moving A Substance
Answer:
(1) 0.333 Hz
(2) 4 sec
(3) 2 sec, 0.5 Hz
Explanation:
(1) We have given time period of pendulum is 3 sec
So T = 3 sec
Frequency will be equal to 
(2) Frequency of the pendulum is given f = 0.25 Hz
Time period is equal to 
(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds
So time taken to complete 1 back and forth swings = 
So time period T = 2 sec
Frequency will be equal to 
