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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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Assignment # 2
allsm [11]

Answer:

1.Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship. The first factor of production is land, but this includes any natural resource used to produce goods and services.

2.Management skills can be defined as certain attributes or abilities that an executive should possess in order to fulfill specific tasks in an organization. They include the capacity to perform executive duties in an organization. ... and practical experience as a manager.                                                                              3.No corrective action is required when the deviations are within acceptable limits. However, when the deviations go beyond the acceptable range, in the important areas, it demands immediate managerial attention so that deviations do not occur again and standards are accomplished.

5 0
3 years ago
one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n
vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

\begin{bmatrix}1 & -1  & 11\\3 & -4 & 4 \end{bmatrix}

\begin{bmatrix}1 & 0  & 40\\0 & 1 & 29 \end{bmatrix}

a = 40

b = 29

6 0
3 years ago
Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system
sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user.  The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 0
3 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
4 years ago
Marks
Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

3 0
3 years ago
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