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3 years ago

Key evidence that mars may once have had much more water than it does today-perhaps enough to form an ocean-comes from ________.

1 answer:

7 0

The answer to this question is precisely the "measurement of the atmospheric ratio of the given deuterium to the given ordinary hydrogen". Hence, if we complete the statement in the problem, we have it, the key evidence that Mars may once have had much more water than it does today- perhaps enough to form or accommodate an ocean comes from the measurement of the atmospheric ratio of the deuterium to the given ordinary hydrogen.

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a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

stiv31 [10]

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

3 0

3 years ago

A longitudinal wave is shown.<br> Which label identifies a rarefaction?

aev [14]

Answer:

Explanation:

9 0

3 years ago

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$

Let u = 0

0.2 = 0×t + $\frac{1}{2}$ × 9.8 × $t^{2}$

$t^{2}$ = $\frac{0.2}{4.9}$

$t^{2}$ = 0.04

t = 0.2s

Now, the launch velocity of the marble launcher can be calculated by:

Speed = Distance / Time

Speed = $\frac{0.73 + 6.3}{0.2}$

Speed = $\frac{6.93}{0.2}$

Speed = 34.65 m/s

Therefore, the launch velocity of the marble launcher is 34.65 m/s

Know more about Launch velocity: -brainly.com/question/18883779

#SPJ9

3 0

1 year ago

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 77.0 kg, is moving to the right at 4.00

Alex777 [14]

Answer:3.31m/s², to the right

Explanation:

According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.

Momentum = mass × velocity

M1 and M2 be the masses of the first and second skaters respectively

Let u1 and u2 be the velocities of the first and second skaters respectively.

v be their common velocity after collision

M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²

According to the law we have

M1u1 + M2u2 = (M1+M2)v

77(4) + 66(2.5) = (77+66)v

308 + 165 = 143v

V = 473/143

V = 3.31m/s²

Their velocity after collision will become 3.31m/s²

They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.

5 0

3 years ago

When making a budget, you should prioritize ___________ before "__________." A. Needs/wants B. Wants/needs C. Electricity/Water

Blizzard [7]

Answer A

Explanation: because u need to make sure you know the difference between a need and a want

7 0

2 years ago