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Rudik [331]
3 years ago
11

Who ever helps can you help me with some other ones?

Physics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

2 seconds

Explanation:

It only went up by 2

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Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to
wlad13 [49]

Answer:

m,lkj,mkn,njn

Explanation:because she is telling you to do a project

7 0
3 years ago
Electricity flows from what to what
DerKrebs [107]
Electricity flows from positive to negative
5 0
2 years ago
Read 2 more answers
Box A has two moles of gas at a temperature of 400 K. Box B has one mole of gas at a temperature of 800 K. Which of the two gas
mina [271]
B I think if not sorry
8 0
3 years ago
Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets
schepotkina [342]
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.
7 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

T^2 = \frac{4\pi^2 d^3}{2}

T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}

T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
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