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Paraphin [41]
3 years ago
10

Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets

?
Physics
1 answer:
schepotkina [342]3 years ago
7 0
I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is:  There's a <em>pair</em> of forces, <em>in both directions</em>, between
every two masses.

-- The strength of the force depends on the <em>product</em> of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal. 
It's the product that counts.  Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers.  More distance => weaker force.  Actually, (more distance)² ==> weaker force.

-- The forces are <em>equal in both directions</em>.  Your weight on Earth is exactly equal to
the Earth's weight on you.  You can prove that.  Turn your bathroom scale face down
and stand on it.  Now it's measuring the force that attracts the Earth toward you. 
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth. 
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal.  But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

--  This works exactly the same for every pair of masses in the universe.  Gravity
is everywhere.  You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity".  It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon 
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons.  (That's
about 2.205 pounds).  The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram.  In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.
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HELP PLZ!!!!!!!!!!!!!!
Sophie [7]

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      <em>Numbers represent seconds</em>

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

7 0
2 years ago
1.
Brilliant_brown [7]

Answer:

1) solvent

2) liquid-solid

3) gas - gas

Explanation:

1) A solution is the homogenous mixture of two or more substances in such a way that its components are uniformly distributed. In any solution, the solvent make up the greatest quantity or volume while the solute is of lesser quantity.

2) liquid-solid is the most common type of solution.

3) A gas solution can be a gas dissolved in a gas, or a liquid dissolved in a gas, or a solid dissolved in a gas.  Air is a gas - gas solution, it is composed of oxygen and other dissolved gases in nitrogen

6 0
2 years ago
Equal amounts of water are kept in a cap and in a dish which will evaporate faster ?why
Ber [7]
The water in a dish evaporates quickly as the surface area of dish is greater than cap which enables the water to evaporate quickly
4 0
3 years ago
Read 2 more answers
A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel
mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

\vec v_{ra} = 12 m/s North

so it will be

Let North direction is along Y axis and East direction is along X axis

\vec v_{ra} = 12\hat j

also it is given that speed of air is 6.7 m/s relative to ground

\vec v_a = 6.7 \hat i

now as we know by the concept of relative motion

\vec v_{ab} = \vec v_a - \vec v_b

\vec v_{ra} = \vec v_r  - \vec v_a

now by rearranging the terms

\vec v_r = \vec v_{ra} + \vec v_a

\vec v_r = 12 \hat j + 6.7 \hat i

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

v_r = \sqrt{12^2 + 6.7^2}

v_r = 13.7 m/s

so the net speed of Robin with respect to ground will be 13.7 m/s

7 0
3 years ago
How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J
algol13

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

Hence, the correct option is (a).

6 0
3 years ago
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