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liubo4ka [24]
3 years ago
14

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

ges could you make that would change the value for |a| to 2.00 m/s2?
Physics
1 answer:
Eddi Din [679]3 years ago
5 0

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Answer:

0.51

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a = 0.973 m/s²

f = kinetic frictional force

Force equation for the motion of the book is given as

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<h3>Explanation</h3>

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Volume of water displaced:

The density of water at room temperature is 1.000\;\text{kg}\cdot\text{dm}^{-3}. Each kilogram of water will occupy a volume of 1 dm³ (one cubic decimeter), which is the same as 1 L (one liter).

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Volume of the object:

The object is completely under water. As a result, the volume of the object will be the same as the volume of water displaced. The volume of the object is also 17.3 L.

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