Answer:
If we put pressure on a solid or a liquid, there is essentially no change in volume. ... The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles.
Explanation:
The technical definition of latitude is the angular distance north or south from the earth's equator measured through 90 degrees. ... Locations at lower latitudes receive stronger and more direct sunlight than locations near the poles. Energy input from the sun is the main driving force in the atmosphere.
The Seasons at Different Latitudes
The seasonal effects are different at different latitudes on Earth. Near the equator, for instance, all seasons are much the same. Every day of the year, the Sun is up half the time, so there are approximately 12 hours of sunshine and 12 hours of night.
When we consider Latitude alone as a control, we know that the low latitudes (say from the Equator to approximately 30 degrees N/S) are the warmest across the year (on an annual basis).
Answer:
rm = 38280860.6[m]
Explanation:
We can solve this problem by using Newton's universal gravitation law.
In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m
![r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]](https://tex.z-dn.net/?f=r_%7Be%7D%20%3D%20distance%20earth%20to%20the%20astronaut%20%5Bm%5D.%5C%5Cr_%7Bm%7D%20%3D%20distance%20moon%20to%20the%20astronaut%20%5Bm%5D%5C%5Cr_%7Bt%7D%20%3D%20total%20distance%20%3D%203.84%2A10%5E8%5Bm%5D)
Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.
Mathematically this equals:

![F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]](https://tex.z-dn.net/?f=F_%7Bm%7D%20%3DG%2A%5Cfrac%7Bm_%7Bm%7D%2Am_%7Ba%7D%20%20%7D%7Br_%7Bm%7D%20%5E%7B2%7D%20%7D%20%5C%5Cwhere%3A%5C%5CG%20%3D%20gravity%20constant%20%3D%206.67%2A10%5E%7B-11%7D%5B%5Cfrac%7BN%2Am%5E%7B2%7D%20%7D%7Bkg%5E%7B2%7D%20%7D%20%5D%20%5C%5Cm_%7Be%7D%3D%20earth%27s%20mass%20%3D%205.98%2A10%5E%7B24%7D%5Bkg%5D%5C%5C%20m_%7Ba%7D%3D%20astronaut%20mass%20%3D%20100%5Bkg%5D%5C%5Cm_%7Bm%7D%3D%20moon%27s%20mass%20%3D%207.36%2A10%5E%7B22%7D%5Bkg%5D)
When we match these equations the masses cancel out as the universal gravitational constant

To solve this equation we have to replace the first equation of related with the distances.

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.
![r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]](https://tex.z-dn.net/?f=r_%7Bm1%2C2%7D%3D%5Cfrac%7B-b%2B-%20%5Csqrt%7Bb%5E%7B2%7D-4%2Aa%2Ac%20%7D%20%20%7D%7B2%2Aa%7D%5C%5C%20%20where%3A%5C%5Ca%3D80.25%5C%5Cb%3D768%2A10%5E%7B6%7D%20%5C%5Cc%20%3D%20-1.47%2A10%5E%7B17%7D%20%5C%5Creplacing%3A%5C%5Cr_%7Bm1%2C2%7D%3D%5Cfrac%7B-768%2A10%5E%7B6%7D%2B-%20%5Csqrt%7B%28768%2A10%5E%7B6%7D%29%5E%7B2%7D-4%2A80.25%2A%28-1.47%2A10%5E%7B17%7D%29%20%7D%20%20%7D%7B2%2A80.25%7D%5C%5C%5C%5Cr_%7Bm1%7D%3D%2038280860.6%5Bm%5D%20%5C%5Cr_%7Bm2%7D%3D-2.97%2A10%5E%7B17%7D%20%5Bm%5D)
We work with positive value
rm = 38280860.6[m] = 38280.86[km]
Answer:
Mass = 18.0 kg
Explanation:
From Hooke's law,
F = ke
where: F is the force, k is the spring constant and e is the extension.
But, F = mg
So that,
mg = ke
On the Earth, let the gravitational force be 10 m/
.
3.0 x 10 = k x 5.0
30 = 5k
⇒ k =
................ 1
On the Moon, the gravitational force is
of that on the Earth.
m x
= k x 5.0
= 5k
⇒ k =
............. 2
Equating 1 and 2, we have;
= 
m = 
= 18.0
m = 18.0 kg
The mass required to produce the same extension on the Moon is 18 kg.
Answer:
...........................