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makkiz [27]
3 years ago
14

Which features on mars point to the possibility of liquid water on the planet?

Physics
2 answers:
Stells [14]3 years ago
7 0
     In 2011 NASA researchers identified features on Mars known as Recuring Slope Lineae ( or RSL ). RSL are relatively dark and narrow features.They are thought to be a signs of salty liquid water on Mars. One hypothesis for RSL formation is that they form when underground bodies of salty water leak into the surface.
   Answer: Recuring Slope Lineae. 
DedPeter [7]3 years ago
5 0

gullies and stream-like channels

Explanation: correct on gradpoint

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Answer:

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3 years ago
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A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

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3 years ago
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Genrish500 [490]

Answer:

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3 years ago
Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

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3 years ago
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geniusboy [140]

Answer:

Explanation:

true

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3 years ago
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