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3 years ago

15

HOW OLD IS THE ISS? HOW LONG HAS IT BEEN OPERATIONAL?

2 answers:

sp2606 [1]3 years ago

5 0

The ISS is 22 years old. The ISS became fully operational in May2009, when it began hosting a six-person crew; this required the constant docking of two Soyuz lifeboats with the ISS. Three Russians, two Americans, and one astronaut from either Japan, Canada, or the European Space Agency made up the six-person crew.

xxMikexx [17]3 years ago

4 0

Answer:

bhai nam kya hai (whats your name)

Explanation:

please comment

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A car weighs 1323 N. what is its mass?

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If the car is on the moon, its mass is about 817 kg.

If it's on the Earth, its mass is about 135 kg.

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How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s squared

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3000 newton force is required

Explanation:

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Which from radiation from the electromagnetic spectrum is MOST LIKELY to cause mutations in growing tissues, and should be MOST

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A coiled spring with coils that are closely spaced then widely then closely then widely then closely, ending with a yellow line

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2, sorry I'm late but my answer is right, I just took the quiz

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During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f

dezoksy [38]

Answer:

$F= 2.3733 x10^{20} N$

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

$M_ {sun} = 1.99 * 10^{30} Kg$

$M_ {Earth} = 5.98 * 10 ^ {24} Kg$

$M_ {Moon} = 7.36 * 10 ^{22} Kg$

Average distances

$\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m$

$\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m$

Gravitational constant

$G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}$

The formula of the Gravitational Force between the Moon and the Earth would be,

$F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}$

$F = 1.9908 * 10 ^{20} N$

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

$F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}$

$F_2 = 4.3641*10^-{ 20} N$

This force is in the direction of the Sun

The net force must be

$F_ {net} = F_2-F$

$F_ {net} = 2.3733*10^{20} N$

This in the direction of the Sun.

8 0

3 years ago