At 2:00 p.m. a car's speedometer reads 30 mi/h. At 2:30 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:30 the
1 answer:
Answer:
The acceleration is exactly 40 mi/h² as shown
Explanation:
Given;
initial velocity of the car, u = 30 mi/h
final velocity of the car, v = 50 mi/h
change in velocity, ΔV = v - u
ΔV = 50 mi/h - 30 mi/h = 20 mi/h
initial time, t₁ = 2:00 PM
final time, t₂ = 2:30 PM
Change in time, Δt = t₂ - t₁
Δt = 2: 30 - 2:00 = 30 mins = 0.5 hour
Acceleration is given as change in velocity per change in time;
a = ΔV / Δt
Therefore, the acceleration is exactly 40 mi/h² as shown.
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Answer:
2.67kg
Explanation:
The maximum velocity, of a body experiencing simple harmonic motion is given by equation (1);
where is the angular velocity and A is the amplitude.
The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);
where k is the force constant of the spring and m is the loaded mass.
We can make the subject of formula in equation (1) as follows;
We then combine equations (2) and (3) as follows;
According to the problem, the following are given;
We then substitute these values into equation (4) and solve for the unknown mass m as follows;
Squaring both sides, we obtain the following;