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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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A world class sprinter can accelerate at 5.0 m/s squared, from the rest in the starting blocks to their top speed in 4.0 seconds
Airida [17]

Answer:

The distance covered by the sprinter, s = 40 m

Explanation:

Given data,  

The initial velocity of the sprinter, u = 0 m/s

The acceleration of the sprinter, a = 5 m/s²

The time period of acceleration of the sprinter, t = 4 s

Using the II equations of motion

                   s = ut + ½ at²

                       = 0 + ½ (5) (4)²

                       = 40 m

Hence, the distance covered by the sprinter, s = 40 m                    

4 0
2 years ago
41. 2072 Set E Q.No. 11 A source of sound produces a note of
Nimfa-mama [501]
A. is the answer for this
4 0
2 years ago
Help it multiple-choice 16 and 17 only
lilavasa [31]

Answer:

16. c

17. d

Explanation:

16. Newton’s third law states that a force will always have an opposite but equal force as a reaction, so every force comes as a pair of action-reaction forces. For example, if you push on a book, the book also pushes on you.

17. If 1cm represents 15N, then 5cm must represent 15N*5=75N.

3 0
2 years ago
A basketball player achieves a hang time of
sp2606 [1]
Time - 0.792s
acceleration - 9.8m/s2
vertical height - ?
remember that height is in meters
so from the acceleration
9.8 = x
----
0.792*0.792
cross multiply
=6.147m approximately 6.2m
8 0
3 years ago
9.
mote1985 [20]

Answer:

D. location

Explanation:

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

That's why all objects regardless of their mass, shape, and size free fall towards the Earth with an acceleration equal to the acceleration at that location in the absence of air resistance.

According to the gravitational force between two bodies, the force experienced by one body due to the other is independent of its own mass.

The gravitational force is given by equation

                                  F =  GMm/r²

If F is the force acting on the smaller body of mass 'm', then

                                 F = ma

Therefore, the equation becomes,

                                ma = GMm/r²

                                   a = GM/r²

The value of 'a' changes with respect to the value of 'r' such that if 'r' is the radius of the Earth, then the acceleration at a height 'h' from Earth surface is given by

                                 a = GM/(r+h)²

Here it is clear that the acceleration at any point is only the inherent property of the Earth itself.

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

6 0
3 years ago
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