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Jlenok [28]
3 years ago
9

a satellite is traveling in a orbit around the earth. what change would cause it to experience more gravitational force from the

earth? A. orbiting closer to the earths surface. B. removing some of its mass. C making it orbit faster. D. making it's surface reflect the sun's rays.
Physics
1 answer:
never [62]3 years ago
7 0
The correct answer is:
<span>A. orbiting closer to the earths surface. 

In fact, the gravitational force exerted by the Earth on the satellite is
</span>F=G \frac{Mm}{r^2}
<span>where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's surface

We can see that, if the satellite orbits closer to the Earth's surface, its distance r from the centre of the planet decreases. But when r decreases, F (the gravitational force) increases, so A is the correct answer.</span>
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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
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Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

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L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
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Answer:

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(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

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We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

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So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

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(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

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