Net Force = (mass) x (acceleration) (Newton #2)
Net Force = (50 kg) x (6 m/s² down)
Net Force = (50 * 6) (kg-m/s² down)
<em>Net Force = 300 Newtons down</em>
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done



The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.


The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy





The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
Answer:





Explanation:
To calculate average velocity we need the position for both instants t0 and t1.
Now we will proceed to calculate all the positions we need:





Replacing these values into the formula for average velocity:




To know the actual velocity, we derive the position and we get:

Explanation:
Newton's second law of motion is F = ma, or force is equal to mass times acceleration