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Mademuasel [1]
3 years ago
6

a motorcycle starting from rest has an acceleration of 2.6m/s how long does it take the motorcycle to travel a distance of 120

Physics
1 answer:
Setler79 [48]3 years ago
5 0
46.15 seconds to reach 120
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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

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

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Is how high or low sound is
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Answer:

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According to Boyle’s Law, increasing the pressure of a gas will have which effect on its volume?
Alisiya [41]
The volume will decrease

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3 years ago
A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul
gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

\theta is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

\theta=90^{\circ}

and the formula reduces to

F=qvB

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

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