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xenn [34]
4 years ago
14

A dragster in a race accelerated from stop to 60 m/s by the time it reached the finish line. The dragster moved in a straight li

ne and traveled from the starting line to the finish line at 8.0 m/s2. How long did it take the dragster?
Physics
1 answer:
Aleks [24]4 years ago
7 0
Vf = final velocity
vo = initial velocity
a = acceleration
t = time

use the following equation

vf = vo + at

since vo = 0 m/s (stopped), that term drops out and you're left with . . .

vf = at

(60 m/s) = (8.0 m/s²)t

t = (60 m/s)/(8.0 m/s²) = 7.5 seconds

<u><em>t = 7.5 seconds</em></u>


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denpristay [2]

Answer:

Jupiter

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5 0
3 years ago
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
garik1379 [7]

(a) 6.04 rev/s

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v=\omega r

where

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r is the distance of the ball from the centre of the circle

In situation 1), we have

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r = 0.900 m

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(b) 1561 m/s^2

The centripetal acceleration of the ball is given by

a=\frac{v^2}{r}

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For situation 1),

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(c) 1292 m/s^2

For situation 2 we have

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a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2

5 0
3 years ago
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sp2606 [1]

Answer:

1.98 rev/s

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r_{f} = final distance of masses in each hand = 0.1 m

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I_{i} = initial total moment of inertia = I + 2 mr_{i}^{2}

w_{i} = initial angular velocity = 1 rev/s

I_{f} = final total moment of inertia = I + 2 mr_{f}^{2}

w_{f} = final angular velocity = ?

Using conservation of angular momentum

I_{i} w_{i} = I_{f} w_{f}

(I + 2 mr_{i}^{2}) w_{i} = (I + 2 mr_{f}^{2}) w_{f}

(5 + 2 (5) (1)^{2}) (1) = (5 + 2 (5) (0.1)^{2}) w_{f}

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3 0
4 years ago
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The correct answer is: Crescent moon.

I hope this helps you!

Brainliest answer is always appreciated!
3 0
3 years ago
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