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lawyer [7]
2 years ago
10

Crews at the International Space Station are researching the effects of the weightlessness of space on ________.

Physics
2 answers:
Natalija [7]2 years ago
8 0

The answer is A. The human body

andrezito [222]2 years ago
6 0

A: Human Body

C is wrong because they don’t have the tools to test it on another planet

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Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
How long have advances in astronomy been occurring? Only for the past few years
solniwko [45]

I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years).  Maybe longer.

3 0
3 years ago
Akia is balancing the equation Na + H2O NaOH + H2. He tries to find the coefficients that will balance the equation. How could h
bixtya [17]

Answer:

by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products. - This is the answer.

Explanation:

3 0
2 years ago
Read 2 more answers
Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s
dedylja [7]
Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.
8 0
3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
2 years ago
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