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Leni [432]
3 years ago
14

Correct answer will receive a Brainliest.

Physics
2 answers:
Gre4nikov [31]3 years ago
8 0
The answer would be c
san4es73 [151]3 years ago
4 0

Answer:

C

Explanation:

The period of a pendulum is found by the equation: T = 2pi*sqrt(L/g). Let the original length be L and the original period be T. The length increased by a factor of 4, so it’s new length is 4L. We get that the new period is 2pi*sqrt(4L/g) = 2pi*2sqrt(L/g) = 4pi*sqrt(L/g). We can see that the period increased by a factor of 2 because the original period, T, equals 2pi*sqrt(L/g) and the new period is 4pi*sqrt(L/g) = 2(2pi*sqrt(L/g)) = 2T. Therefore, the new period is 2(1.4) = 2.8

I hope this helps! :)

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Convection is the heat transfer due to the bulk movement of molecules within fluids such as gases and liquids, including molten rock. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. These clouds may produce precipitation, which is the primary route for water to return to the Earth's surface within the water cycle.

Explanation:

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4 years ago
12.What did Chargaff said about DNA molecules?
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Answer:

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3 years ago
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Jobisdone [24]

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5 0
3 years ago
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Alexxandr [17]

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3 0
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

7 0
3 years ago
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