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andre [41]
2 years ago
5

Two objects have the same center point of the circle, but are located at different positions away from the center point. Each ob

ject is moving with uniform circular motion.
Which would describe the tangential speed of the objects?

both objects would have the same tangential speed
the object with the smaller radius has a faster tangential speed
the object with the larger radius has a faster tangential speed
both objects would have oscillating tangential speeds
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

The object with the larger radius has a faster tangential speed. Tangential speed is related to both rotational speed and radial distance from the rotating axis.

<h3>What is uniform circular motion?</h3>

Uniform circular motion is a type of motion of a particle around a circle at a constant speed. The magnitude of the speed of the particle is constant.While the direction is changing continuously.

Tangential speed is related to both rotational speed and radial distance from the rotating axis.

The object with the larger radius has a faster tangential speed.

Hence, option C is correct.

To learn more about the uniform circular motion, refer to the link;

brainly.com/question/2285236

#SPJ1

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Because of the different speeds..

7 0
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9. You drive a car from Milwaukee to Chicago, which is a distance of 150km and it takes you 95
Katena32 [7]

For this case we have to, by definition:

v = \frac {x} {t}

Where:

v: It's the velocity

x: It is the distance traveled

t: It is the time spent

1 hour equals 60 minutes and 1 minute equals 60 seconds.

On the other hand:

1 kilometer equals 1000 meters

So:\frac {150} {95} \frac{km} {min} * \frac {1} {60} \frac {min} {sec} * \frac {1000} {1} \frac {m} {km} = \frac {150 * 1000} {95 * 60} \frac {m} {sec} = \frac {15000} {5700} \frac {m} {sec} = 2.63 \frac {m} {sec}

On the other hand:\frac {150} {95} \frac{km} {min} * \frac {60} {1} \frac {min} {hr} = \frac {150 * 60} {95} \frac {km} {h} = 94.74 \frac {km} {h}

ANswer:

2.63 \frac {m} {sec}\\94.74 \frac {km} {h}

8 0
4 years ago
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how
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Answer:

W = 354.9 J

Explanation:

Given that,

The mass of a bale of hay, m = 23 kg

The displacement, d = 3.9 m

The horizontal force exerted on the hay, F = 91 N

We need to find the work done. We know that,

We know that,

Work done, W = Fd

So,

W = 91 N × 3.9 m

W = 354.9 J

So, the required work done is 354.9 J.

4 0
3 years ago
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