Answer:
a) 
b) 
c) 
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:

Part b
For this case first we can convert the spring constant to N/m like this:

And the work donde by the spring on this case is given by:

Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

And if we solve for the initial velocity we got:

Part d
Let d1 represent the new maximum distance, in order to find it we know that :

And replacing we got:

And we can put the terms like this:

If we multiply all the equation by 2 we got:

Now we can replace the values and we got:


And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
Answer:
what r the questions i can’t see them
Explanation:
Chaff has also written about his book of Mormon that is not true to say this but it doesn't have the
Explanation:
Take F=ma
a = F/m
For a higher, F higher or m lower
Means higher horse power for engine or lower mass for the car
Hello There!
It takes the planet Mars around 24 hours, 37 minutes, 23 seconds to rotate on its axis. This is around the same amount of time that it takes our planet to rotate once on its axis.