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blondinia [14]
3 years ago
5

What are the variables of Boyle's law?

Physics
1 answer:
telo118 [61]3 years ago
4 0
Pressure and number of molecules of gas,volume and number of molecules of gas.
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A scientist hypothesizes that the temperature at which an turtle's egg is incubated
just olya [345]

Answer:

thw temperature of the male will be higher than that of the female.

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3 years ago
How many planets are in our Solar system?<br> I'll give brainliest
Dima020 [189]

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The following is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune

That is from our solar system

Explanation:

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2 years ago
What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s
jarptica [38.1K]
You would probably have a low frequency due to how much the wavelength is spread out.
8 0
3 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
3 years ago
Two people were pulling this thing with 560 Newton. Now they are using a rope with 50°. How much Newton would they need to pull
Brums [2.3K]

Answer:

10 N

Explanation:

While many people would like to simply add the forces from each end to get a total force, this is fundamentally incorrect.

MIGHT BE TOTALLY WRONG

4 0
3 years ago
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