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Mademuasel [1]
3 years ago
15

A woman lifts a 300 newton child a distance of 1.5 meters in 0.75 seconds. What is her power output in lifting the child?

Physics
1 answer:
Nitella [24]3 years ago
7 0
Power = Work done / time

Work done = Force * Distance
                 =  300 N *  1.5 m =  450 J

Power =  450 / 0.75 = 600 Watts.
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Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere
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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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8 0
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What will happen two temperatures if you increase particle motion
nalin [4]

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
Read 2 more answers
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
3 years ago
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