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dusya [7]
3 years ago
14

If every action has an equal and opposite reaction, then when you jump, and gravity pulls you back down, it also pulls the Earth

up to you. Why can’t you see this motion?
Physics
1 answer:
LenKa [72]3 years ago
8 0

When two objects are gravitationally attracted, the forces

on both of them are equal. As the objects move straight

toward each other, then, just as we'd expect, the object

with the greater mass has the smaller acceleration and

moves a smaller distance by the time they collide, while

the object with a smaller mass has the greater acceleration

and moves a greater distance by the time they collide.


The Earth's mass is roughly approximately something like 70,000,000,000,000,000,000,000 times as much as YOUR mass,

so the difference in acceleration and distance moved before

the collision is really substantial.

You might be interested in
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLE
Romashka [77]

A_x = 5.0

A_y = -6.3

6 0
2 years ago
20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat
uysha [10]

100000 Pascal

Explanation:

pressure= force/area

Max pressure= force/min area

so f=5

min area= 5×10^-5

5÷5*10^-5 = 100000pascal

8 0
2 years ago
Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?
Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0
2 years ago
Read 2 more answers
How to calculate moments with 3 separate weights of different amounts at different points?
Rina8888 [55]
I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it.  But calculating the moment is easy, and we
can get along without the drawing.

Each separate weight has a 'moment'.
The moment of each weight is: 

             (the weight of it) x (its distance from the pivot/fulcrum) .

That's all there is to a 'moment'.

The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).

That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do.  The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.
6 0
3 years ago
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