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Vladimir [108]
3 years ago
7

Which of the following statements is/are true?

Physics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

C and E

Explanation:

I think that is the answers

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From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:
den301095 [7]

Answer: Jupiter's mass

Explanation:

From Kepler's third law:

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

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.

If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3

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The current through a heater is 12 A when it is plugged into a
jasenka [17]

Answer:

A. 10Ω

Explanation:

7 0
2 years ago
A type O star is likely to appear _____. yellow, blue, green or red
Murljashka [212]
A type O star is likely to appear blue. 
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3 years ago
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Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
3 years ago
the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.
asambeis [7]
The\ area\ of\ a\ circle\ =  \pi r^2 \\ 725\ cm^2 =  3.14r^2 \\
230.57 cm^2 = r^2 \\ \sqrt{230.57\ cm^2} = r \\ \boxed{r = 15.18\  cm}
7 0
2 years ago
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