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svetoff [14.1K]
3 years ago
5

At what distance from the Earth’s surface is a 10,000 kg satellite if its potential energy is equal to –5.58 x 1011 J? (choose t

he closest answer)
A 958 km

B 858.km

C 758 km

D 658 km
Engineering
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

Option D

Explanation:

As we know

Gravitational Constant (G) = 6.67*10^{11}

Mass of Earth (M) = 5.98*10^{24}

Weight of satellite (m) = 10,000 Kg

Radius of Earth (R) = 6.378*10^6

Distance of satellite from Earth = \frac{GMm}{R} - U

Substituting the given values, we get

D=\frac{(-((6.67*10^{11})(5.98*10^{24})(10000))}{(6.378*10^6)} - (-5.58*10^{11})\\\\0.625*10^{3}Km

or 635 m

The correct answer is option D

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Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process

Explanation:

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Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans
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Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

Diameter of the tube, D = 10 cm = 10\times 10^{- 2} m = 0.1 m

Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, we make use of the steady flow eqn:

h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

Specific Volume at outlet, V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s

From the eqn:

P_{o}V_{o} = mRT_{o}

m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s

Now,

p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

Also,

\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

4 0
3 years ago
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
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Answer:

± 0.003 ft

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Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

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alina1380 [7]

Answer: True

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Answer:

Explanation:

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3 years ago
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