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saul85 [17]
3 years ago
8

Help needed Can a body be accelerating if it is moving in circle

Physics
2 answers:
kupik [55]3 years ago
7 0

Answer:

an object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction. Because the speed is constant for such a motion, many students have the misconception that there is no acceleration. "After all," they might say, "if I were driving a car in a circle at a constant speed of 20 mi/hr, then the speed is neither decreasing nor increasing; therefore there must not be an acceleration." At the center of this common student misconception is the wrong belief that acceleration has to do with speed and not with velocity. But the fact is that an accelerating object is an object that is changing its velocity. And since velocity is a vector that has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity. For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

Explanation: hopes it help u

belka [17]3 years ago
5 0

<em>Answer:</em>

<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>

<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>

<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>

<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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Answer:

Mechanical advantage = 3

Explanation:

You exert a 100-N force on a pulley system to lift 300-N.

The mechanical advantage of the system is given by the ratio of output force to the input force.

Here, output force = 300 N and input force = 100 N

Mechanical advantage,

m=\dfrac{300}{100}\\\\m=3

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The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun
Kipish [7]

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

T=2\pi \sqrt{\frac{l}{g}}         (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}

g': acceleration due to gravity at the top of the mountain

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g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

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