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taurus [48]
4 years ago
14

A wheel with rotational inertia i is mounted on a fixed, frictionless axle. the angular speed ω of the wheel is increased from z

ero to ωf in a time interval t. what is the average net torque τ on the wheel during this time interval?
Physics
1 answer:
Annette [7]4 years ago
4 0
According to Newton's second law:
τ = Iα, where I is the moment of inertia and α is the angular acceleration.
Angular acceleration can be written as:
α = (ωf - ωi) / t
α = ωf/t

Substituting into the second law:
τ = Iωf/t
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D. has no overall force acting on it.

Explanation:

Why?

Because in a straight line at the constant speed means the car moving in the same velocity, which is not acceleration neither deceleration, and it cannot be on a downhill slope. So the correct answer is

<h3>→ D. has no overall force acting on it.</h3>
8 0
3 years ago
Describe when the chemical<br> reaction occurs in a dry-cell<br> battery
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Answer:

<em>For a dry-cell battery to operate, oxidation will occur from the zinc anode and reduction will take place in the cathode. The most common type of cathode is a carbon graphite. Once reactants have been turned into products, the dry-cell battery will work to produce electricity.</em>

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8 0
2 years ago
15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m
Rudik [331]

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7 0
3 years ago
Read 2 more answers
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
A student walks 31 m east, then 16 m west. Make east positive. What is their<br> displacement?
Marianna [84]
Displacement = 31 - 16 = +15 m
3 0
3 years ago
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