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3 years ago

7

A machinist turns the power on to a grinding wheel, at rest at tome t=0 s. the wheel accelerates uniformly for 10 s and reaches

the operating angular velocity of 58 rad/s. the wheel is run at the angular velocity for 30 s and then a brake is applied which immediately stops the wheel. how many revolutions does the wheel undergo from t=0 until is stops?

1 answer:

Assoli18 [71]3 years ago

3 0

To convert 2030 rad into rev, divide 2030 by 2pie. So final answer will be 2030/2 pie =323.08 revolutions.

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Answer:

2000 kg m/s

Explanation:

The momentum of an object is a vector quantity whose magnitude is given by

$p=mv$

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m is the mass of the object

v is the velocity of the object

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In this problem, we have:

- Spaceship 1 has

m = 200 kg (mass)

v = 0 m/s (zero velocity)

So its momentum is

$p_1 =(200)(0)=0$

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m = 200 kg (mass)

v = 10 m/s (velocity)

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Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

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We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

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Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

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Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

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Using equation of motion

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Put the value in the equation

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Using formula of distance

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Put the value into the formula

$H'=2397.4-1800$

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