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3 years ago

Objects 1 and 2 attract each other with a gravitational force

Physics

2 answers:

Vinvika [58]3 years ago

7 0

Answer: The new force will be 72.0 units.

Explanation: The gravitational force between two objects can be written as:

F = G*m1*m2/r^2

Where G is the gravitational constant, m1 and m2 are the masses of both objects and r is the distance between them.

Now, we have that F = 18.0 units.

Now, supose that we change r to r/2, the new force is:

F' = G*m1*m2/(r/2)^2 = 4*G*m1*m2/r^2 = 4*F = 4*18.0 units = 72.0 units

Ivanshal [37]3 years ago

3 0

Answer:

27unit

Explanation:

18×3/2= 9×3= 27unit

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5 N acting in the same direction as the 10 N force

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4 years ago

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

The work is equal to

W = 5886 [J]

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

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4 years ago

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