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timofeeve [1]
3 years ago
10

Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 4.50 ✕ 10

3 kg, and the second a mass of 7.50 ✕ 103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.
Physics
1 answer:
vovangra [49]3 years ago
6 0
Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.
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Almost immeasurably small.

Explanation:

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Let's look at the reports and draw our own conclusions.

Current CO₂ levels are 409.8 parts per million (PPM)

at the beginning of the Industrial revolution in the 1700's, the presumed beginning of the huge increase in CO₂ the level was about 280 PPM

For perspective lets assume we capture the whole atmosphere and squish it down to 2400 one liter bottles of air

That's 100 cases of 24 bottles per case.

We now separate all the air components into their own bottles

Nitrogen is 78% of our air, so we subtract 78 cases from our 100 leaving 22

Subtracting Oxygen at 21% of air leaves 1 case of liter bottles left

Of those 24 bottles, Argon makes up 0.93% of air so we subtract 22 bottles

The remaining two bottles contain all of the other gasses in our air, One of those bottles contains CO₂.

If we take the CO₂ levels from the 1700's at about 280 PPM as a baseline and assume ALL of the increase is human caused, that is (410 -280) / 280 = 46 % of the total.

The human caused addition of CO₂ would be 460 mililiters out of 2400 liters over the course of 250 years 

The claim is, that less than half of a liter of CO₂ out of 2400 liters of air is responsible for heating not only the gas in all the other bottles but also the surface of the earth itself.

Personally, it boggles my mind.

And it says NOTHING of a far more powerful greenhouse gas that is far more prevalent in the atmosphere...water vapor.

Water vapor is about 1% of air at sea level and about 0.4% overall. It was not considered in the above analysis because water vapor can condense out and is not a constant in the air.

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2.99 m/s

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v^2-u^2 = 2as

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0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s

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tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

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Hope this helps you

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