The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:
where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:
where the negative sign indicates that the direction is backward.
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The kinetic energy of the small ball before the collision is
KE = (1/2) (mass) (speed)²
= (1/2) (2 kg) (1.5 m/s)
= (1 kg) (2.25 m²/s²)
= 2.25 joules.
Now is a good time to review the Law of Conservation of Energy:
Energy is never created or destroyed.
If it seems that some energy disappeared,
it actually had to go somewhere.
And if it seems like some energy magically appeared,
it actually had to come from somewhere.
The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.
Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
B. The object stops from a velocity of 12.0 m/s
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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