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Sholpan [36]
3 years ago
13

Which layer of the atmosphere is the top layer of the thermosphere?

Physics
2 answers:
joja [24]3 years ago
8 0
The thermosphere is a layer of Earth's atmosphere. The thermosphere is directly above the mesosphere and below the exosphere. It extends from about 90 km (56 miles) to between 500 and 1,000 km (311 to 621 miles) above our planet.
Westkost [7]3 years ago
3 0
The exosphere is the outermost layer<span> of Earth's </span><span>atmosphere.</span>
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Please help on this one?
Lelu [443]

The easiest way to answer this question is just to get the answer first. The answer is A with the added comment that no chemical reaction has taken place.

Layered means that the chemicals are not soluble in one another. B is not the answer.

C is eliminated by what what was said about A.

D a solution is not a pure substance (singular) by itself. There are at least 2 chemicals together.

6 0
2 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
2 years ago
How far will a 70N crate is move with 3500J of work
r-ruslan [8.4K]
Work = Force * distance
Force = 70 N
Work = 3500 J

3500 = 70d
d = 3500/70 = 50 m
6 0
3 years ago
Read 2 more answers
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
Which vehicle will have more kinetic energy, a parked<br> semitruck or a car moving at 50 km/h?
Neporo4naja [7]

Answer:

a semi truck

Explanation:

3 0
2 years ago
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