Answer:
5.59 m/s
Explanation:
We are given;
Mass = 110 kg
Initial velocity: u = 13.41 m/s
Force = 615 N
Time(t) = 1 s
Now, the formula for force is;
Force = mass x acceleration
Thus;
615 = 110 × acceleration
\Acceleration(a) = 615/110 = 5.591 m/s²
Now, using Newton's first law of motion, we can find acceleration (a). Thus;
v = u + at
v = 13.41 + (5.591 × 1)
v ≈ 19 m/s
So,the change in velocity is;
Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
Answer:

Explanation:
The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

The initial speed of Ryan is:

20.3 divided by 3.0 will get u velocity and v times 3.0s
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