A particle travels in a circle of radius 14 m at a constant speed of 21 m/s. What is the magnitude of the acceleration (in m/s2)
1 answer:
The magnitude of the acceleration is equal to 31.5
<u>Given the following data:</u>
- Radius = 14 meters.
- Speed = 21 m/s
To determine the magnitude of the acceleration:
In Science, the acceleration of a particle in a <u>circle</u> or circular track is known as centripetal acceleration.
<h3>How to calculate centripetal acceleration.</h3>Mathematically, centripetal acceleration is given by this formula:
Substituting the given parameters into the formula, we have;
Centripetal acceleration = 31.5
Read more on centripetal acceleration here: brainly.com/question/25780931
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B) The object's velocity doubled.
Explanation:
The graph is missing: find it in attachment.
The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as
where
m is the mass of the object
v is the velocity of the object
We notice that:
- The kinetic energy is directly proportional to the mass
- The kinetic energy is proportional to the square of the velocity
In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.
First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.
Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore
B) The object's velocity doubled.
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