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Firlakuza [10]
3 years ago
11

Which of the following are known to exist on the moon? select all that apply.

Physics
2 answers:
denis-greek [22]3 years ago
3 0
B. ice and d basalt............
katrin [286]3 years ago
3 0

Answer:

Ice and Basalt.

B. and D.

Explanation:

In 2018, it was confirmed by NASA that water ice exists on the surface of the moon.

In 1972, lunar basalt was picked by astronauts Eugene Cernan and Harrison Schmitt who were on the Apollo 17 mission.

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QUESTION 1 Linear Motion
Misha Larkins [42]

Answer:

48.51ms / 174.6 km/h

Explanation:

y = 1/2 x g x t^2         v = g x t          

when y = 120m

120 = 1/2 x 9.8 x t^2

t^2 = 24.49

t = 4.95s

when t = 4.95s

v = 9.8 x 4.95

v = 48.51 m/s = 174.6 km/h

I'd say its realistic. But I don't really know that sry

3 0
2 years ago
A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
pantera1 [17]

Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

time taken by the wave to reach the top is given as

t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

t = 12 s

4 0
3 years ago
If two people are running at the same speed in the same direction. One person is one meter ahead of the other. The person in fro
Gre4nikov [31]

Answer:yes

Explanation:

5 0
3 years ago
Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi
Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0
3 years ago
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Vitek1552 [10]

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

5 0
3 years ago
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