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3 years ago

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) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

stead had mass 4m, its orbital period would be A) T. B) 2T. C) 4T. D) T/4. E) T/2

1 answer:

Mrrafil [7]3 years ago

4 0

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

$T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

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Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a curre

Igoryamba

Answer:

a) $\frac{F_1}{L}=1.95*10^-^5N$

b) $\frac{F_2}{L}=1.95*10^-^5N$

Explanation:

From the question we are told that:

Distance between wires $d=32.2$

Wire 1 current $I_1=2.75$

Wire 2 current $I_2=4.33$

a)

Generally the equation for Force on $l_1$ due to $I_2$ is mathematically given by

$F_1=I_1B_2L$

Where

B_2=Magnetic field current by $I_2$

$B_2=\frac{\mu *i_2}{2\pi d}$

Therefore

$F_1=I_1B_2L$

$F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L$

$\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }$

$\frac{F_1}{L}=1.95*10^-^5N$

b)

Generally the equation for Force on $I_2$ due to $I_1$ is mathematically given by

$F_2=I_2B_1L$

Where

B_1=Magnetic field current by $I_2$

$B_1=\frac{\mu *I_1}{2\pi d}$

Therefore

$\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})$

$\frac{F_2}{L}=1.95*10^-^5N$

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D !.!.!.!.!.!!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!!..!!..!

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Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

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=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

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How to find a planet’s gravitational field strength using its radius?

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The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

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Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

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$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

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Thermal Expansion. best describes the increase in a materials volume

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