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Genrish500 [490]
3 years ago
7

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security

agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking 13.0 s. What is the ratio of the man's running speed to the sidewalk's speed?
Physics
2 answers:
Zigmanuir [339]3 years ago
6 0
13:9 because he starts back then runs slower
snow_tiger [21]3 years ago
6 0

Answer:

The ratio of the man's running speed to the sidewalk's speed: 5: 26.

Explanation:

Let the distance between the sidewalk from one end to the other be x

Distance covered by the man = x

Time taken by the person to cover x distance = 2.50 seconds

Speed of the man while running to other end point = S

S=\frac{x}{2.50 s}..[1]

Distance covered by the man by running back to his starting point after agents appears = x

Time taken by the person to cover x distance after agents appears = 13.0 seconds

Speed of the man while running back to starting point after agents appears = S'

S'=\frac{x}{13.0 s}...[2]

The ratio of the man's running speed to the sidewalk's speed:

[2] ÷ [1]

\frac{S'}{S}=\frac{\frac{x}{13.0 s}}{\frac{x}{2.50 s}}=\frac{5}{26}

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A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
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Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

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v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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