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Mice21 [21]
3 years ago
8

Two satellites A and B are orbiting the earth around the equator, E, at different altitudes on a circular path of 2400km and 300

0km from the surface of the earth. The radius of the earth at the equator is its equatorial radius which is 6378km, mass of the earth is 5.98*10^24kg, g=9.81m/s^2, G=m^2/kg^2. If the satellites are of the same mass, 100kg, determine the:
a) Force required to keep each satellite in orbit.
b) Speed of the satellite and which is at a higher speed around the earth.
c) The period and the angular frequency of the satellites around the earth.
d) Force of attraction between the earth and each satellite, fA and fB.
Engineering
1 answer:
White raven [17]3 years ago
7 0
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A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
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Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

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3 years ago
Discuss 7 habits of highly effective people and how important are ethics in today's society​
kompoz [17]

Answer:

Explanation:

The 7 Habits of Highly Effective People, is a book written and first published in 1989. It is a business and self-help book that was written by Stephen Covey. The seven habits include

Being proactive

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First things first

Always thinking towards a win-win situation

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7 0
3 years ago
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe
Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}

Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t

Therefore, we have;

Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0
2 years ago
A digital filter is given by the following difference equationy[n] = x[n] − x[n − 2] −1/4y[n − 2].(a) Find the transfer function
slega [8]

Answer:

y(z) = x(z) - x(z) {z}^{ - 2}  -  \frac{1}{4} y(z) {z}^{ - 2}  \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\  h(z) = \frac{y(z)}{x(z)}  =  \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}

The rest is straightforward...

6 0
2 years ago
Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the
alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is 123.788\,\frac{kg}{min}.

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

\dot Q_{ref} - \dot Q_{w} = 0

\dot Q_{ref} = \dot Q_{w}

\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})

The mass flow rate of the cooling water is now cleared:

\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}

Given that h_{ref,in} = 808.34\,\frac{kJ}{kg}, h_{ref, out} = 88.82\,\frac{kJ}{kg}, h_{w,out} = 104.83\,\frac{kJ}{kg} and h_{w,in} = 62.98\,\frac{kJ}{kg}, the mass flow of the cooling water is:

\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)

\dot m_{w} = 123.788\,\frac{kg}{min}

The mass flow rate of cooling water required to cool the refrigerant is 123.788\,\frac{kg}{min}.

4 0
3 years ago
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