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AveGali [126]
4 years ago
6

An astronaut on a spacewalk accidentally drops a tool, and it floats away.Which of the following objects is exerting a gravitati

onal force on the floating tool? Choose all that apply
the astronaut
the Moon
the Earth
the Sun
Physics
2 answers:
castortr0y [4]4 years ago
4 0

Answer:

All of them are correct

Explanation:

They are all correct select all

a_sh-v [17]4 years ago
3 0

Gravitational force is given as

F = \frac{Gm_1m_2}{r^2}

so here gravitational force will be given by above formula where the force between two objects depends on the mass of two objects and distance between them.

Now here since the tool is at finite distance from Astronaut and Moon, Sun, Earth

So all the above will exert gravitational force on the tool but the magnitude of force will be different as all of the above are of different masses and situated at different distance.

So all options are correct here

The Astronaut

The Moon

The Sun

The Earth

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(A) What is the maximum tension possible in a 1.00-millimeter-diameter nylon tennis racket string?
attashe74 [19]

Complete Question

(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?

(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?

The  tensile  strength of the nylon string is  600*10^{6} \  N/m^2

Answer:

A

   T = 471.3 \  N

B

 To prevent breakage the thickness of the string is increased

  String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension

Explanation:

From the question we are told that

     The  diameter is  d =  1.00 \ mm  =  0.001 \  m  

       The  tensile strength of the nylon string is \sigma =  600 *10^{6} \  N/m^2

  Generally the radius is mathematically evaluated as

     r=  \frac{d}{2}

=>    r =  \frac{0.001}{2}

=>     r =  0.0005 \  m

The cross sectional area is mathematically represented as

     A = \pi  r^2

=>   A =  3.142  *  (0.005)^2

=>    A =  7.855*10^{-7}\  m^2

Generally the tensile strength of nylon is mathematically represented as

      \sigma  = \frac{T}{ A }

Where T is the tension on the maximum tension on the string

 So  

           T =  \sigma  *  A

=>          T =  600*10^{6} *  7.855*10^{-7}

=>         T = 471.3 \  N

Form the equation above  we  see that

        T  \  \alpha \  A

So if the tension is  increased to prevent breakage the thickness of the string is increased(i. e the cross-sectional  area )

String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension

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