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OLEGan [10]
2 years ago
14

Describe the relationship between the masses of two objects and the force of gravity between them.

Physics
2 answers:
professor190 [17]2 years ago
7 0

Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.

MakcuM [25]2 years ago
3 0
Because the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. Therefore as the mass of either object increases, the force of gravitational attraction between them also increases.
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Tyler's favourite number is 8642, and he tries to fit it into everything he does. If he wishes to do that much work by climbing
Brums [2.3K]

Answer:

2014

Explanation:

none

6 0
3 years ago
A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

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a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

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3 0
3 years ago
Read 2 more answers
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
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Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
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Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
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What one part of a circuit might cause your calculator to be off?
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Resistors tell me if im right

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