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sashaice [31]
3 years ago
13

The European Space Agency launched a probe called Rosetta in March 2004. In August​ 2014, Rosetta reached its​ destination: a co

met called​ 67P/Churyumov-Gerasimenko. Rosetta is the first spacecraft to rendezvous with a comet. It contained a lander​ craft, called Philae.The Philae has a launch mass on the Earth of 220​pounds-mass ​[lbm​].If the Philae was deployed to Mars instead of Comet​ 67P, what would the weight of the lander be after landing on the

Engineering
1 answer:
Art [367]3 years ago
6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The of the lander would be 75.66 pounds

Explanation:

From the question we are given that the

Weight of the Philae on Earth  is = 220 pound-mass

gravity of Mars is = 3.71 m/s^{2}

Weight of the Philae on Mars is = ?

Now Mass is quantity of matter in an object so it is always constant for a particular object

Generally Weight = Mass × Gravity :

                      \frac{Weight Of Earth}{Gravity Of Earth} = \frac{Weight Of Mars}{Gravity Of Mars}

Hence  

         Weight of the Philae on Mars is = \frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}

                                                              =\frac{200 *3.71}{9.81}

                                                              = 75.66 pounds

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Nata [24]

Answer:

abrir candados y abrir puertas

Explanation:

4 0
2 years ago
(TCO 1) Name one disadvantage of fixed-configuration switches over modular switches. a. Ease of management b. Port security b. F
frutty [35]

Answer:

The answer is A.

Explanation:

Two main types of network switches, modular and fixed configuration switches, are used for connecting the devices with one another provided they are on the same network.

As the name suggests, modular switches can be configured according to your needs and specific situations where you need a different setup.

The one advantage fixed-configuration switches have over the modular switches is that they are easier to operate. You can't change anything for a different application but they are simpler to setup and use, you can just plug them in and start using. They are usually for the more casual end-user and home networks etc.

I hope this answer helps.

5 0
3 years ago
One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp
pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}

1x0.723x(T_{eq} -350)=3x0.780x(450-T_{eq} ) ⇒T_{eq} = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

V_{a,i}  = \frac{mRT_{a,i} }{P_{a,i} }=  \frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar} = 0.201m^{3}

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }

P_{eq}= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

                             (0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0
3 years ago
On diesel engines, data from ________ sensors are commonly used to adjust exhaust gas recirculation (EGR) rates.
e-lub [12.9K]

Answer:

Air mass sensors is the right answer i think

Explanation:

3 0
2 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

3 0
3 years ago
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