Answer:
a) -1.25 m/s²
b) 62.5 m
Explanation:
Convert km/h to m/s:
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
a = Δv / Δt
a = (0 m/s − 12.5 m/s) / 10 s
a = -1.25 m/s²
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 12.5 m/s) (10 s)
Δx = 62.5 m
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Answer:
The net power needed to change the speed of the vehicle is 275,000 W
Explanation:
Given;
mass of the sport vehicle, m = 1600 kg
initial velocity of the vehicle, u = 15 m/s
final velocity of the vehicle, v = 40 m/s
time of motion, t = 4 s
The force needed to change the speed of the sport vehicle;
The net power needed to change the speed of the vehicle is calculated as;
Answer:
ms⁻¹
Explanation:
= diameter of merry-go-round = 4 m
= radius of merry-go-round = = = 2 m
= moment of inertia = 500 kgm²
= angular velocity of merry-go-round before ryan jumps = 2.0 rad/s
= angular velocity of merry-go-round after ryan jumps = 0 rad/s
= velocity of ryan before jumping onto the merry-go-round
= mass of ryan = 70 kg
Using conservation of angular momentum
ms⁻¹
Answer:
The time where the avergae speed equals the instaneous speed is T/2
Explanation:
The velocity of the car is:
v(t) = v0 + at
Where v0 is the initial speed and a is the constant acceleration.
Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:
v_ave = v0+a(T/2)
We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>
v(T/2)=v_ave
Now we want to know where the car should be, the osition of the car is:
Where x_A is the position of point A. Therefore, the car will be at:
<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>