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kozerog [31]
3 years ago
7

If F1 is the magnitude of the force exerted

Physics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer: 3. F1 = F2

Explanation:

According to <u>Newton's law of Gravitation</u>, the force F exerted <u>between two bodies</u> or objects of masses M and m and separated by a distance r is equal to the product of their masses divided by the square of the distance:  

F=G\frac{Mm}{r^2} (1)

Where Gis the gravitational constant

Now, in the especific case of the Earth and the satellite, where the Earth has a mass M and satellite a mass m, being both separated a distance r, the force exerted  by the Earth on the satellite is:

F1=G\frac{Mm}{r^2}   (2)

And the force  exerted by the satellite on the Earth is:

F2=G\frac{Mm}{r^2}   (3)

As we can see equations (2) and (3) are equal, hence the magnitude of the gravitational force is the same for both:

F1=F2

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Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

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196 hp = 146157 W

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Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

P=-F\times v

F=\dfrac{-P}{v}

F=\dfrac{-146157\ W}{32\ m/s}

F = -4567.40 N

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3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

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             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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Answer:

The ratio of the diameter of iron to Cu is;

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Explanation:

R=(ρL)/A

  • R is resistance,
  • L is length,
  • A is area,
  • ρ is resistivity
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from the question the two materials have the same resistance per unit length.

\frac{R}{L}= \frac{p}{A}

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This means we can equate ρ/A for both materials.

\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }

re-arranging the equation we have,

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A=\pi \frac{d^{2} }{4}

\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe}   }{ d^{2}{Cu}   }

\frac{d^{2}{Fe}   }{ d^{2}{Cu}   } =\frac{p_{Fe} }{ p_{Cu} }

\frac{d{Fe}   }{ d{Cu}   } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}

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Answer: This is the orbit (of the moon around Earth).

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