Answer:
a) V = 0.82m/s
b) Vmax = 0.985 m/s
Explanation:
By conservation of energy we know that:
Eo = Ef 
Solving for V we get:
V = 0.82 m/s
To find the maximum speed we will do the same to an intermediate point where the compression is X and the distance for the work donde by frictions is given by (Xmax - X) = (0.28m - X):

Then we have to solve for V, derive and equal zero in order to find position X. After solving the derivative we get:
X = 0.1m Replacing this value into the equation for Vmax:
Vmax = 0.985m/s
Answer:
The earth, The sun, the solar system and the milky way.
The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
The question can be solved, using Newton's second law of motion.
Note: Momentum of the cannon = momentum of the cannonball.
<h3>
Formula:</h3>
- MV = mv................. Equation 1
<h3>Where:</h3>
- M = mass of the cannon
- m = mass of the cannonball
- V = velocity of the cannon
- v = velocity of the cannonball
Make v the subject of the equation.
- v = MV/m................ Equation 2
From the question,
<h3>Given: </h3>
- M = 500 kg
- V = 3 m/s
- m = 10 kg.
Substitute these values into equation 2.
- v = (500×3)/10
- v = 150 m/s.
Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
Learn more about Newton's second law here: brainly.com/question/25545050
Answer:
E = 58.7 V/m
Explanation:
As we know that flux linked with the coil is given as

here we have


now we have

now the induced EMF is rate of change in magnetic flux

now for induced electric field in the coil is linked with the EMF as





Answer:I have to say 56
Explanation: because it is going up by 8