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velikii [3]
3 years ago
14

g The tires of a car make 75 revolutions as the car reduces its speed uniformly from 91 km/h to 48 km/h . The tires have a diame

ter of 0.78 m . Part A What was the angular acceleration of the tires
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

   α = -0.01625 rad / s²

Explanation:

This is an exercise in angular kinematics, we can use the relation

         w = w₀ + 2 α θ

linear and angular variables are related

        v = w r

         w = v / r

Let's reduce the magnitudes to the SI system

         v₀ = 91 km / h (1000m / 1km) (1h / 3600s) = 25.278 m / s

         v = 48 km / h = 13,333 m / s

         θ  = 75 rev (2π rad / 1 rev) = 471.24 rad

Let's find the angular velocities

         w₀ = v₀ / r

         w₀ = 25.278 / 0.78

         w₀ = 32,408 rad / s

         w = v / r

         w = 13.333 / 0.78

         w = 17.09 rad / s

we calculate the angular acceleration

         α = (w- w₀) / 2θ  

         α = (17.09 - 32.408) / (2 471.24)

         α = -0.01625 rad / s²

the negative sign indicates that the wheel is stopping

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Sauron [17]

Answer:

7.08 m/s²

Explanation:

Given:

v₀ = 20.0 m/s

v = 105 m/s

t = 12.0 s

Find: a

v = at + v₀

105 m/s = a (12.0 s) + 20.0 m/s

a = 7.08 m/s²

5 0
3 years ago
Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped
Akimi4 [234]

Answer: F = 102141N

Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

a=\frac{\Delta v}{\Delta t}

a=\frac{11}{0.14}

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>

6 0
3 years ago
In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
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<span>Kinetic energy increases and potential energy decreases.
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3 years ago
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If your apparatus were to be dropped from a mile above the ground, describe the forces acting upon your apparatus as it fell. Ho
kvv77 [185]

Answer:

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Accelerometers have many uses in industry and science. Highly sensitive accelerometers are used in inertial navigation systems for aircraft and missiles. Vibration in rotating machines is monitored by accelerometers. They are used in tablet computers and digital cameras so that images on screens are always displayed upright. In unmanned aerial vehicles, accelerometers help to stabilise flight.

When two or more accelerometers are coordinated with one another, they can measure differences in proper acceleration, particularly gravity, over their separation in space—that is, the gradient of the gravitational field. Gravity gradiometry is useful because absolute gravity is a weak effect and depends on the local density of the Earth, which is quite variable.

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Explanation:

hope this helps !!!!

7 0
2 years ago
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