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3 years ago

Uncle Fester's CD's

Physics

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MrRissso [65]3 years ago

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Answer: hey

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What is Newton’s first law

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Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

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3 years ago

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Were the continents once joined together as a supercontinent? Give 3 pieces of evidence to support Alfred Wegeners Theory of Con

sattari [20]

Yes! Fossils, The outlines of the continents and geological features .

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3 years ago

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What is the work done if you push a tree with<br> 50 N of force but the tree does not move?

Soloha48 [4]

Answer:

The work done is 0.

Explanation:

The reason no work is done is because the equation W = Fs.

W = work

F= force

s= displacement

In this scenario F = 50 and s= 0

Therefore.

W = 50(0)

W = 0

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3 years ago

A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off

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No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

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3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

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

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

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

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

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3 years ago