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3 years ago

Answer this question please and you will can’t the brain list

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

We know that velocity equals to difference in distance divided by difference in time.

1st period: velocity is increasing. Where v = (8-0)/(3-0) = 8/3 m/s.

2nd one: velocity decreases. Where

v=(5-8)/(4-3) = (-3) m/s.

3rd one: velocity is constant because no change in distance, so v=0.

4th one: velocity increases again. Where v=(10-5)/(8-6)= 5/2 m/s.

5th one: velocity is constant again, so v=0.

6th one: velocity decreases, where

v=(6-10)/(15-12)= (-4)/3 m/s.

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The net torce on an object moving with constant speed in circular motion is in which direction?

aleksley [76]

The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

6 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

By the term universe astronomers mean

OverLord2011 [107]

'Universe' means 'Everything'. That is, all matter, all space, all time.

6 0

4 years ago

Please answer this question I have uploaded the picture of question please answer

8090 [49]

Answer:

20 mangintiude beacuse

Explanation:

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3 years ago

Describe the steps used by conventional sewage treatment plants to break down our sewage into a relatively safe effluent that ge

anyanavicka [17]

Answer:

the soild material is allowe to settle into a sludge layer the remaining waste water undergoes a secondart treatment that uses bacteria to break down organic matter

treat water is left to sit while particles settle out, with settle d particles being added to sluudge,

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Explanation:

5 0

3 years ago