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Firdavs [7]
3 years ago
7

The Imager for Mars Pathfinder (IMP) is an imaging system. It has two camera channels. Each channel has color capability. This i

s provided by a set of filters. It was developed by a team led by the University of Arizona. Scientists and engineers from around the world were a part of the team. The system has three parts: (1) It has a camera head. (2) It has an extendable mast with electronic cabling. (3) It has two plug-in electronic cards. Each card plugs into a slot within the spacecraft.
What is the main idea of the numerical list in the second paragraph?
A. the electronic cabling in the imaging system
B. the three parts of the imaging system
C. the camera head of the imaging system
D. the two electronic cards in the imaging system
E. the extendable part of the imaging system
Engineering
1 answer:
Crank3 years ago
5 0
I believe it’s E. Not completely shure though
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2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
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Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

<u>Determine the force transmitted by the coupling between the nozzle and hose </u>

attached below is the remaining part of the  detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2  

= 32 ( 25 / 35 )^2

= 16.33 m/s

4 0
2 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

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3 years ago
How to update android 4.4.2 to 5.1​
faust18 [17]

Answer:

try settings and go to updates?

Explanation:

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Answer:

eh I'm good hbu?????????

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4. Partnership programs between schools and the owners
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Answer:

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Explanation:

Basically hiring students for hands on training to learn the basics of mechanics.

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