The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
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And.. where is the rest of the question?
Answer:
The velocity of the ship relative to the earth V = 9.05 
Explanation:
The local ocean current is = 1.52 m/s
Direction 
= 40°
Velocity component in X - direction 
= 1.52 
°
= 1.164
Velocity component in Y - direction 
= 8 + 1.52 
°
= 8.97
The velocity of the ship relative to the earth
Put the values of and we get,
⇒ 
⇒ V = 9.05
This is the velocity of the ship relative to the earth.
Answer:
11.78meters
Explanation:
Given data
Mass m = 100kg
Length of cord= 10m
Spring constant k= 35N/m
At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy
That is
Us=Ug
Us= 1/2kx^2
Ug= mgh
1/2kx^2= mgh
0.5*35*10^2= 100*9.81*h
0.5*35*100=981h
1750=981h
h= 1750/981
h= 1.78
Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge
