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astraxan [27]
3 years ago
13

Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's sur

face, while that for satellite b is at a height of 838 km. find the orbital speed for satellite a and satellite
b.
Physics
1 answer:
Evgen [1.6K]3 years ago
6 0
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite 
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g' 
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h): 
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)] 
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec 
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
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3 years ago
Two 4.3546 cm x 4.3546 cm square aluminum electrodes, spaced 0.6408 mm apart are connected to a 73.68 V battery. What is the cap
Helen [10]

Answer:

Capacitance, C = 26.1 picofarad

Explanation:

It is given that,

Side of square, x = 4.3546 cm = 0.043546 m

Distance between electrodes, d = 0.6408 mm = 0.0006408 m

Voltage, V = 73.68 V

Capacitance of parallel plates is given by :

C=\dfrac{\epsilon_oA}{d}

C=\dfrac{8.85\times 10^{-12}\times (0.043546)^2}{0.0006408}

C=2.61\times 10^{-11}\ F

or

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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
kakasveta [241]

Answer:

W=4037.36\ J

Explanation:

Given:

mass of ice melted, m=3.3\times10^{-2}\ kg

time taken by the ice to melt, t=5\ min=300\ s

latent heat of the ice, L=3.34\times 10^5\ J

Now the heat rejected by the Carnot engine:

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Q_R=0.033\times 3.34\times 10^5

Q_R=11022\ J

Since we have boiling water as hot reservoir so:

T_H=373\ K

The cold reservoir is ice, so:

T_L=273\ K

Now the efficiency:

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\eta=1-\frac{273}{373}

\eta=26.81\%

Now form the law of energy conservation:

Heat supplied:

Q_S-W=Q_R

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Q_S= heat supplied to the engine

Q_S-\eta\times Q_S=Q_R

Q_S(1-\eta)=Q_R

Q_S=\frac{11022}{1-0.2681}

Q_S=15059.36\ J

Now the work done:

W=Q_S-Q_R

W=15059.36-11022

W=4037.36\ J

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