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Tatiana [17]
3 years ago
5

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ng the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 4.1 m/s, and 65-kg person feels a 455-N force pressing against his back. What is the radius of a chamber?
Physics
1 answer:
ozzi3 years ago
4 0

Answer:

r=2.4m

Explanation:

We have to use the centripetal force  equation

Fc=\frac{mv^{2} }{r}

we  need the radious so we have to isolate "r" and we get

r=\frac{mv^{2} }{Fc}

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

r=\frac{65*4.1^{2} }{455}

r=2.4m

The radius of the amusement park chamber is 2.4m

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Answer:

Force (P) : Positive

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Weight (mg) : Zero

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The work done by a force on an object is given by the following formula:

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W = Work Done

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<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

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W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

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<u></u>

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3 years ago
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8 0
2 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

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1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

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Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
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