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Nata [24]
3 years ago
6

Which change will always result in an increase in the gravitational force between two objects?

Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

Increasing the masses of the objects and decreasing the distance between the objects.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

F_g=\frac{GmM}{d^2}

Where, 'G' is called the universal gravitational constant.

Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects and directly proportional to the masses of the two objects.

This means that, if masses increase with other parameters constant, force will increase and vice-versa. If distance decreases with other parameters constant, force will increase and vice-versa.

So, on combining the above two relations,we are certain that if one increases the masses and decrease the distance between, the gravitational force will definitely increase.

So, the third option is correct.

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

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<em>t</em><em>h</em><em>e</em><em> </em><em>l</em><em>e</em><em>n</em><em>g</em><em>t</em><em>h</em><em> </em><em>i</em><em>s</em><em> </em><em>m</em><em>e</em><em>a</em><em>s</em><em>u</em><em>r</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>b</em><em>e</em><em> </em><em>2</em><em>s</em>

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