Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, 
Force applied on the tracks, 
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
So, the required time is 300.3 seconds.
PE stands for Potential Energy. It is the stored energy in an object due to its position with respect to some reference. It is expressed in Joules.
P.E = m * g * h OR P.E. = mgh
m - mass of the body
g - acceleration due to gravity
h - height attained due to the body's displacement.
K.E. stands for Kinetic Energy. It is the energy possessed by a body due to its motion.
K.E = 1/2 mv² where m = mass of the body; v = velocity with which the body is moving.
The correct answer is:
~A. The greater the distance to a galaxy, the greater its redshift.
Hope this helps!!!
