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andreyandreev [35.5K]
3 years ago
5

Which tools collect images during space exploration? Check all that apply. rovers orbiters satellites space shuttles space stati

ons
Physics
2 answers:
riadik2000 [5.3K]3 years ago
6 0

Answer:

rovers

satellites

space stations

Explanation:

i just took the test

grigory [225]3 years ago
4 0

Answer:

1. Rovers- These are vehicles which are designed to move on the surface on any celestial body and collect samples, data and images. Sometimes, these can be used to transport mission crew members.

2. Orbiters- These are spacecrafts designed to orbit any celestial body and collect data in form of images.

3. satellites- artificial satellites are placed in the orbit of celestial bodies for various purposes. There are various types of satellites like communication satellite, weather satellites, space telescopes. These collect images and data inform of signals.

4. space stations: It is a satellite where research and experiments take place. Images are also collected for research purpose.

Space shuttle is a wrong option because it is just a vehicle to transport these satellites and probes into the space and place in the orbit or surface of celestial body.

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A train leaves the station heading south on the tracks. It takes 5 seconds to reach 50 miles per hour. It completes the entire 1
Musya8 [376]
Well, in order to figure out the answer is to divide until you figure out how many miles they went per second. If it takes 5 seconds to reach 50 miles per hour it took 10 seconds per every 10 miles meaning each mile took 1 second. (Not actually possible but the answer) So, If it finished a 100 mile trip in 2 hours it took an hour for 50 miles. If it took 1 hour for 50 miles divide 60/50 which gets you 1.2 so it took 1.2 miles per minute meaning the car went 120 miles per hour I believe. I hope this helps :)
8 0
3 years ago
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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

468 kg m/s + 0 = (137 kg)v

v = \dfrac{468\  kg m/s }{137 \ kg}

v = 3.416 m/s

8 0
3 years ago
Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in
monitta

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>

  • Initial velocity=u=0m/s
  • Final velocity=v=10m/s
  • Time=10s=t

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}

\\ \sf\longmapsto Acceleration=\dfrac{10}{10}

\\ \sf\longmapsto Acceleration=1m/s^2

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>

  • initial velocity=0m/s=u
  • Final velocity=v=0.25m/s
  • Time=t=2s

\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}

\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}

\\ \sf\longmapsto Acceleration=0.125m/s^2

Person-1 is accelerating faster.

4 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg
nata0808 [166]

Answer:

a) U_{e} = 3 \times 10^{10}\,J, b) v \approx 7745.967\,\frac{m}{s}

Explanation:

a) The potential energy is:

U_{e} = Q \cdot \Delta V

U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)

U_{e} = 3 \times 10^{10}\,J

b) Maximum final speed:

U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }

The final speed is:

v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }

v \approx 7745.967\,\frac{m}{s}

3 0
3 years ago
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The answer is B which is exactly what happens.

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3 years ago
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