a) For the motion of car with uniform velocity we have ,
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting

The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting

Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209 
Work is force times distance. If there's no distance, there's no work being done.
Answer:
61.33 Kg
Explanation:
From the question given above, the following data were obtained:
Distance = 1×10² m
Time = 9.5 s
Kinetic energy (KE) = 3.40×10³ J
Mass (m) =?
Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:
Distance = 1×10² m
Time = 9.5 s
Velocity =?
Velocity = Distance / time
Velocity = 1×10² / 9.5
Velocity = 10.53 m/s
Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:
Kinetic energy (KE) = 3.40×10³ J
Velocity (v) = 10.53 m/s
Mass (m) =?
KE = ½mv²
3.40×10³ = ½ × m × 10.53²
3.40×10³ = ½ × m × 110.8809
3.40×10³ = m × 55.44045
Divide both side by 55.44045
m = 3.40×10³ / 55.44045
m = 61.33 Kg
Thus, the mass of Leroy Burrell is 61.33 Kg
Answer:
The gravitational potential energy of a squirrel is 53.312 J.
Explanation:
We have,
Mass of a squirrel is 0.68 kg
It is placed at a height of 8 m above the ground.
It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

So, the gravitational potential energy of a squirrel is 53.312 J.
i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.
hope this helps!