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2 years ago

Can someone pls help I’m stuck!

Physics

1 answer:

Elan Coil [88]2 years ago

6 0

B because why not i was wondering how ppl are so smart and let me tell ya i don’t know

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

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2 years ago

Explain how buoyancy can defeat the force of gravity

Olenka [21]

Answer:

Buoyancy causes things to float in water or liquid which appears to defeat the force of gravity. Hope this helps.

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2 years ago

Radioactive isotopes can be used to find the age of rocks, fossils, or other artifacts. Carbon-14 has a half-life of 5,730 years

bazaltina [42]

Answer:

1/8 = (1/2)^3

This implies the sample has decayed for 3 half lives

3 * 5730 yrs = 17,200 years

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1 year ago

What is the surface area of an object weighing 100M which exerts a pressure of 20nm-2

vovangra [49]

lovely question hope 7 solve it

Explanation:

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2 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

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

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

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

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

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2 years ago