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3 years ago

14

PLZZZZZ HELP ASAP I WILL MARK BRAINIEST FOR WHOEVER HAS THE BEST ANSWER!!!!!! THIS IS MY THIRD TIME PUTING THIS QUESTION UP I DO

NT WANT TO DO IT AGAIN!!!
What are some criteria and constraints for sending a rover to Venus?
I only need 2 criteria and 2 constraints
THANK YOU SO MUCH TO THE PEOPLE THAT WILL HELP MEE

1 answer:

ddd [48]3 years ago

8 0

Answer:

Well first for criteria think what would the rover need in order to sustain itself on Venus. And for constraints think of anything that could possibly affect the rover( ex: gasses, active volcanoes)

Explanation:

Criteria: Make the rover self sustainable, and allow the rover to have a mission on Venus( ex: collect rock samples)

Constraints, as I mentioned above gasses, and active volcanoes.

I hope this helps! :)

You might be interested in

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350MV . Th

AleksAgata [21]

Complete Question

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350 MV (35,000,000 V). The bottoms of the thunderclouds are typically 1500 m above the earth, and can have an area of 120 km^2. Modeling the earth/cloud system as a huge capacitor, calculate

a. the capacitance of the earth-cloud system

b. the charge stored in the "capacitor"

c. the energy stored in the capacitor

Answer:

a

$C = 7.08 *10^{-7} \ F$

b

$Q = 24.78 \ C$

c

$E =433650000 \ J$

Explanation:

From the question we are told that

The potential difference is $V = 35000000 V$

The distance of the bottom of the thunderstorm from the earth is d = 1500 m

The area is $A = 120 \ km^2 = 120 *10^{6} \ m^2$

Generally the capacitance of the earth cloud system is mathematically represented as

$C = \epsilon_o * \frac{A}{d}$

Here $\epsilon_o$ is the permitivity of free space with as value $\epsilon_o = 8.85 *10^{-12} \ C/(V\cdot m)$

So

$C = 8.85*10^{-12} * \frac{120*10^{6}}{1500}$

=> $C = 7.08 *10^{-7} \ F$

Generally the charge stored in the capacitor (earth-cloud system) is mathematically represented as

$Q = C * V$

=> $Q = 7.08 *10^{-7} * 35000000$

=> $Q = 24.78 \ C$

Generally the energy stored in the capacitor is mathematically represented as

$E = \frac{1}{2} * Q * V$

=> $E = \frac{1}{2} * 24.78 * 35000000$

=> $E =433650000 \ J$

8 0

3 years ago

An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q

Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency $f = ( \dfrac{1}{2 \pi}) \omega$

where;

$\omega = \sqrt{\dfrac{k}{m} }$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )$

However;

$k = \dfrac{F}{x}$ and;

mass $m = m_{car } + m_{person}$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )$

where;

$F = m_{person}g$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )$

replacing the values;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )$

$\mathbf{f = 0.442 \ Hz}$

8 0

3 years ago

List some things in your house that has the same density....

natali 33 [55]

Same Density as what??

4 0

4 years ago

PLEASE HELP ME Color corresponds to the ______________ of light waves. wave speed cycles wavelength

lorasvet [3.4K]

Wavelength. Each wavelength is a certain color. For instance, shorter wavelengths (like 470nm) will be blue or violet, while longer wavelengths (like 650nm) will be red. Hope this helps! :)

5 0

3 years ago

When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a

Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}$

$\dfrac{1}{f}=-\dfrac{3}{110}$

$f=-36.6\ cm$

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

$\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}$

Here, focal length is positive for converging lens

$\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}$

$\dfrac{1}{v}=\dfrac{367}{20130}$

$v=54.85\ cm$

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0

3 years ago