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Neko [114]
2 years ago
7

Which of the following describes a polar orbit?

Engineering
1 answer:
Aleksandr-060686 [28]2 years ago
4 0
Low altitude is the answer
You might be interested in
A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h
anygoal [31]

Answer: Attached below is the missing diagram

answer :

A)   1) Wr > WI,     2) Qc' > Qc

B)   1) QH' > QH,   2) Qc' > Qc

Explanation:

  л = w / QH = 1 - Qc / QH  and  QH = w + Qc

<u>A) each cycle receives same amount of energy by heat transfer</u>

<u>(</u> Given that ; Л1 = 1/3 ЛR )

<em>1) develops greater bet work </em>

WR develops greater work ( i.e. Wr > WI )

<em>2) discharges greater energy by heat transfer</em>

 Qc' > Qc

solution attached below

<u>B) If Each cycle develops the same net work </u>

<em>1) Receives greater net energy by heat transfer from hot reservoir</em>

QH' > QH   ( solution is attached below )

<em>2) discharges greater energy  by heat transfer to the cold reservoir</em>

Qc' > Qc

solution attached below

4 0
2 years ago
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br
vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation v = v_o + a_ct to determine the velocity of car B.

where;

v_o = initial velocity

a_c = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

S = d + v_ot + \dfrac{1}{2}at^2

Then:

v_B = 60-12t

The distance traveled by car B in the given time (t) is expressed as:

S_B = d + 60 t - \dfrac{1}{2}(12t^2)

For car A, the needed time (t) to come to rest is:

v_A = 60 - 18(t-0.75)

Also, the distance traveled by car A in the given time (t) is expressed as:

S_A = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

Relating both velocities:

v_B = v_A

60-12t = 60 - 18(t-0.75)

60-12t =73.5 - 18t

60- 73.5 = - 18t+ 12t

-13.5 =-6t

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

S_B = S_A

d + 60 t - \dfrac{1}{2}(12t^2) = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60  * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0
3 years ago
In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.
egoroff_w [7]

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.

4 0
2 years ago
How are industries related to systems
REY [17]

Answer:

The industrial systems of the future are seen as complex systems, composed of vast numbers of devices, interacting with each other and with enterprise systems continuously.

summary:

they are related because they ARE a system. well, a type.

hope this helps!! :)

7 0
2 years ago
What is the uppermost part of the tree referred to as?
tangare [24]

Answer:

canopy I believe it is called

7 0
2 years ago
Read 2 more answers
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