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otez555 [7]
3 years ago
14

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

l to 90.0 N is applied to the rim of the wheel. The wheel has radius 0.150 m. Starting from rest, the wheel has an angular speed of 14.3 rev/s after 2.72 s. What is the moment of inertia of the wheel
Physics
1 answer:
maksim [4K]3 years ago
3 0

Answer:

the moment of inertia of the wheel is 0.408 kg.m²

Explanation:

Given;

tangential force applied to the wheel, f = 90 N

radius of the wheel, r = 0.15 m

initial angular speed of the wheel, ω₁ = 0

final angular speed of the wheel, ω₂ = 14.3 rev/s

time of motion of the wheel, t = 2.72 s

The tangential acceleration of the wheel is calculated as;

\alpha _t = \alpha r

where;

\alpha is the angular acceleration

\alpha = \frac{\Delta \omega }{\Delta t} = \frac{\omega _2 - \omega_1}{t_2-t_1}\\\\where;\\\\\omega_2 \ is \ the \ final \ angular \ speed = 14.3 \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 89.86 \ rad/s

\alpha = \frac{89.86 - 0}{2.72} = 33.04 \ rad/s^2\\\\\alpha _t = \alpha r\\\\\alpha _t = 33.04 \ rad/s^2 \times 0.15 \ m\\\\\alpha _t = 4.96 \ m/s^2

The mass of the wheel is calculated as;

F = ma

m = F/a

m = (90)/(4.96)

m = 18.15 kg

The moment of inertia of the wheel is calculated as;

I = mr²

I = 18.15 x (0.15)²

I = 0.408 kg.m²

Therefore, the moment of inertia of the wheel is 0.408 kg.m²

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. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
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8 0
3 years ago
What is the acceleration of a 0.30-kg volleyball when a player uses a force of 42 N to spike the ball?
quester [9]

Answer:

The acceleration will be 140 meter per second

Explanation:

Force F = mass m × acceleration a

If F = 42 N and m = 0.30 kg

Then  acceleration a = F/m

a = 42/0.30

a = 140 m/s

5 0
4 years ago
Select the correct answer.
aleksklad [387]
A. Alternating current
4 0
3 years ago
How am I supposed to solve this?
RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

7 0
3 years ago
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