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1 answer:
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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.
Answer:
The moment of inertia of large ring is 2MR².
(A) is correct option.
Explanation:
Given that,
Mass of ring = M
Radius of ring = R
Moment of inertia of a thin ring = MR²
Moment of inertia :
Moment of inertia is the product of the mass of the ring and square of radius of the ring.
We need to calculate the moment of inertia of large ring
Using formula of moment of inertia
Where, = moment of inertia at center of mass
M = mass of ring
R = radius of ring
Put the value into the formula
Hence, The moment of inertia of large ring is 2MR².