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3 years ago

Where might water be found on the moon

Physics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Water Could be Found in the frozen Ice on the moon. it would most likely be underground however

NNADVOKAT [17]3 years ago

4 0

I think water would be found in the middle of the moon aka the core of the moon. If there was water on the moon it could either be found in rocks or in the core. The reason i said core instead of water is because the moon doesn't really have rocks like Mars does it's sandy or ashy on the moon.

Hope this helped :)

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Identify global climate zones and characteristics of each

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2 years ago

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3 years ago

510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr

Yuki888 [10]

Answer:

The terminal velocity is $v_t =17.5 \ m/s$

Explanation:

From the question we are told that

The mass of the squirrel is $m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg$

The surface area is $A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2$

The height of fall is h =4.8 m

The length of the prism is $l = 23.2 = 0.232 \ m$

The width of the prism is $w = 11.6 = 0.116 \ m$

The terminal velocity is mathematically represented as

$v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }$

Where $\rho$ is the density of a rectangular prism with a constant values of $\rho = 1.21 \ kg/m^3$

$C$ is the drag coefficient for a horizontal skydiver with a value = 1

A is the area of the prism the squirrel is assumed to be which is mathematically represented as

$A = 0.116 * 0.232$

$A = 0.026912 \ m^2$

substituting values

$v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }$

$v_t =17.5 \ m/s$

7 0

3 years ago

9. Which of the following statements is true about scientific theories?

AleksAgata [21]

I think the correct answer from the choices listed above is option D. Scientific theories summarize patterns found in nature. Although, the statement scientific theories are never proven is somewhat true. They are either disproved or they are never disproved. Hope this answers the question.

6 0

3 years ago

Read 2 more answers

A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res

VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

$KE = PE$

$\frac{1}{2} mv^2 = mgh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(2(55.8*10^{-2}))}$

$v = 4.67m/s$

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

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3 years ago