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3 years ago

14

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

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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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago

What is refraction?

Damm [24]

B- light bends as it passes through an object ( a would be reflect)

4 0

3 years ago

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$

In the y axis

$v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}$

The resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s$

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

4 0

3 years ago

A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM

ra1l [238]

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0 => Fe = m*g

Using Hook's law:

K*X = m*g Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m

3 0

3 years ago

Read 2 more answers

2) If the initial velocity of an object is equal to a final velocity, what is the acceleration of the object?

Arlecino [84]

Answer:

The acceleration is 0.

Explanation:

The formula for acceleration is:

a = (v-u)/t

If v and u are equal, thus it would be 0 when subtracted, and anything divided by 0 is 0.

8 0

3 years ago