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kumpel [21]
4 years ago
12

What does the force of gravity between two objects depend on?

Physics
1 answer:
lawyer [7]4 years ago
3 0

-- The equal, mutual forces of gravity between two objects depend on the product
of the two masses ... it doesn't matter whether the masses are equal or there's
one huge mass and one minuscule mass. It's the product that counts.

-- The force also depends on the distance between the centers of the two objects.


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Dmitry_Shevchenko [17]

Answer:

deceleration is the opposite of acceleration

Explanation:

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2 years ago
A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
3 years ago
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2/25/20 or 2/28/20 Dispatch #53
mixer [17]

Answer:

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Explanation:

Initial data:

F = 35[N]

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In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

Power = 21 [W]

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3 years ago
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Answer:

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Explanation:

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Black_prince [1.1K]
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