Answer:
Mass of the vehicle and small bug.
Explanation:
- By Newton's third law, force on bug and vehicle will be same when they collide with each other irrespective of their masses.
- But according to Newton's second law, force is mass times acceleration. Since the force on each mass is same, the smaller mass will accelerate more and the heavier mass will accelerate less for the same force.
- Therefore the acceleration of bug will be very greater than vehicle as the mass of the bug is very small as compared to vehicle.
Learn more about Newton's law.
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If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.
Let N be the normal force that forces the person against the wall.
Then u N = m g is the frictional force supporting the person's weight
and N = m g / u
also, N = m v^2 / R is the normal force providing the centripetal acceleration
So, m g / u = m v^2 / R
v^2 = g R / u
since v = 2 pi R T
4 pi^2 R^2 T^2 = g R / u and T^2 = g / (4 u pi^2 R)
T = 1/ (2 pi) (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2
T = .68 / s
Do you see any thing wrong here?
T should have units of seconds not 1 / seconds
v should be 2 * pi * R / T where T is the time for 1 revolution
So you need to make that correction in the above formula for v.
This question involves the use of the equations of motion for vertical motion.
The time taken by the stones P and Q to reach the ground is the same, that is "2 s".
The velocity with which Q hits the ground is "20 m/s".
The time taken by the stones to reach the ground can be calculated by using the second equation of motion for the vertical motion:

For both the stones P and Q:
h = height = 20 m
= initial velocity = 0 m/s
t = time = ?
g = acceleration due to gravity = 10 m/s²
Therefore,

<u>t = 2 s</u>
<u></u>
Hence, the time taken by both the stones to reach the ground <u>is the same</u>.
To find the final velocity of stone Q we will use the first equation of motion for the vertical motion:

Learn more about equations of motion here:
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The attached picture shows the equations of motion in the horizontal and vertical directions.