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3 years ago

Do all planetary systems look the same as our own?

Physics

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

Explanation:

A planetary system consists of at least one star and non stellar objects revolving around it.

Our solar system has one star around which there are 8 planets. However there are star systems with more than one star. These systems are called binary systems. The size of stars also vary. They also vary by orbital configuration i.e, the planets have higher eccentricity than our solar system's. The planetary systems are also classified on the basis of the number of planets in them.

So, all planetary systems do not look the same as our own.

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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

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3 years ago

According to Newton's first law, which characteristic of a moving object would remain constant if there were no other

suter [353]

In object in motion stays in motion; speed

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3 years ago

Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds

vivado [14]

Strong alien you got there good luck bud you never asked a question

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3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$