Please any one need help on this
1 answer:
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Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²
Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ =
τ =
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω =
ω =
ω =
A = 0.52 rad/s B = -0.182 rad/s²