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gladu [14]
4 years ago
9

A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. I

f this object is moved 500 km farther from the planet, the gravitational force on it will be closest to A) 71 N. B) 25 N. C) 56 N. D) 50 N. E) 75 N.
Physics
1 answer:
Mars2501 [29]4 years ago
7 0
<h2>Option C is the correct answer.</h2>

Explanation:

Gravitational force is given by

                  F=\frac{GMm}{r^2}

         G=6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²

         M = Mass of object 1

          m = mass of object 2

          r = Distance between objects.

Here only variable is r value.

In case 1

             100=\frac{GMm}{(1000+500)^2}\\\\GMm=100\times 1500^2

In case 2

            F=\frac{GMm}{(1000+500+500)^2}\\\\F=\frac{GMm}{2000^2}\\\\F=\frac{100\times 1500^2}{2000^2}\\\\F=56.25N

Option C is the correct answer.

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A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  \vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s

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      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

Here the negative sign in \frac{dx}{dt} is because of the negative sign that is attached to x in the equation

 So

       \frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]

From the given equation of velocity field

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    v_y  =  0.2t

     v_z  =  -1.4

So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

substituting the given values of x and t

\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]]      

\frac{dT}{dt} = -0.8 +0.84 + 0.976  

\frac{dT}{dt} = 1.016 ^oC/m  

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