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
Power used by the clock=1.03 W
Explanation:
resistance= 14000 ohm
voltage=120 V
The formula for the power is given by

P=(120)²/14000
P=1.03 W
Answer:
The velocity of the fish hitting the ground is , v = 45.795 m/s
Explanation:
Given data,
The mass of the fish, m = 5 kg
The height of the bird from the surface, h = 107 m
Using the III equation of motion,
v² = u² + 2gs
<em> v = √(u² + 2gs)</em>
Substituting the values,
v = √(0² + 2 x 9.8 x 107)
= 45.795 m/s
Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s
Answer:

Explanation:
Given that,
The mass of a Hubble Space Telescope, 
It orbits the Earth at an altitude of 
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

Where
is the mass of Earth
Put all the values,

So, the potential energy of the telescope is
.