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pishuonlain [190]
3 years ago
15

A race finishes at the same place as it started. The contestants start from rest and cross the finish line running. Which one of

the following is true about the average speed, the average velocity, and the average acceleration? Note that the symbol ≠ means "not equal to".
a). average speed = 0 m/s, average velocity ≠ 0 m/s, average acceleration = 0 m/s2
b).average speed ≠ 0 m/s, average velocity = 0 m/s, average acceleration ≠ 0 m/s2
c).average speed = 0 m/s, average velocity = 0 m/s, average acceleration = 0 m/s2
d).average speed ≠ 0 m/s, average velocity = 0 m/s, average acceleration = 0 m/s2
e.)average speed ≠ 0 m/s, average velocity ≠ 0 m/s, average acceleration =0m/s^2
Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

b).average speed ≠ 0 m/s, average velocity = 0 m/s, average acceleration ≠ 0 m/s2

Explanation:

When a costentant starts a race, he/she acquire speed, velocity and acceleration.

Speed is just a quantity that describes position changing over time. Velocity is similar to speed, but here we have to take on account direction because velocity is a vectorial unit. To describe velocity of a body you must have the speed plus the direction the body is moving at a certain direction.

On the other hand, acceleration is the rate at which velocity changes over time. Like speed, is only a quantity and does not need a direction to be fully described.

So, in this problem, the constestants start from rest and finish running in the same point, It means that at the end the contestants' speed is diferent from 0. Then, the average velocity of a contestant, because of its vectorial nature, and because the contestant starts and finish in the same point, the sum of all velocity vectors throughout the race is equall to 0.

Finally, assuming the contestant accelerates at the beginning (starts with speed = 0 m/s and runs up to speed = x m/s )and keeps a stable speed throughout the race, and because it finishes running, the average acceleration is different from 0.

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To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
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                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
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                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

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                    N_y= (40*9.8)-(169.8*sin59)=246.4N

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                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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