Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
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Weak winds that blow for short periods of time with a short fetch.
Answer:
Torque, 
Explanation:
It is given that,
Length of the wrench, l = 0.5 m
Force acting on the wrench, F = 80 N
The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :



So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.
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