A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is
1 answer:
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Answer:
Explanation:
Given
Magnitude of charge is
Force experienced is
Electric field intensity is the electrostatic force per unit charge
Thus, the electric field intensity is
Answer:
(a)
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So
(a) We know that force is given by
F = ma
So force will be
(b) From second equation of motion we know that
We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as
(d) Time t = 1.5 sec
So