A 5-gram [g] rubber ball is released from a height of 8 meters [m] above a flat surface on the moon. Gravitational accelerati
on on the moon is 1.62 meters per second squared [m/s2]. Assume that no energy is lost from frictional drag. What is the velocity, in units of meters per second [m/s], of the rubber ball the instant before it strikes the flat surface?
2 answers:
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Answer:
Explanation:
a)
Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.
100 x 10cm = 1000
Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....
x 1000
= 1000g or 1kg
Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....
1000g + 100g = 1100g or 1.1kg
b)
The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....
20g ÷ = 2.5
c)
Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 overflowed. So now we the same process as in number a) just with a few adjustments.
x (1000
- 2.5
) = 997.5g
So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.
100g + 997.5g + 20g = 1117.5g or 1.1175kg
Answer:
106.7 N
Explanation:
We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:
where
F is the average force
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem:
m = 0.200 kg
u = 20.0 m/s
v = -12.0 m/s
Solving for F,
And since we are interested in the magnitude only,
F = 106.7 N