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2 years ago

Which layer of the atmoshere is the top layer of the thermoshere

Physics

2 answers:

Sloan [31]2 years ago

8 0

exosphere is the outer layer of the atmosphere

d1i1m1o1n [39]2 years ago

7 0

Answer:

The answer is Exosphere.

Explanation:

Firstly, a brief denition of thermosphere:

The thermosphere is a layer of Earth's atmosphere. The thermosphere is directly above the mesosphere and below the exosphere. It extends from about 80 km to between 600 and 800 km above our planet.

The thermosphere is typically about 200° C (360° F) hotter in the daytime than at night, and roughly 500° C (900° F) hotter when the Sun is very active than at other times.

I have attached an image that shows the atmosphere layers

Finally, the top layer of the thermosphere is Exosphere.

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A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever

Amanda [17]

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

4 0

3 years ago

Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th

Shkiper50 [21]

The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is rod B

Reason:

The elongation of a rod by the application of a force is given by the

following formula;

$\Delta L = \dfrac{F \cdot L}{A \cdot E}$

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

a given force reduces as the cross sectional area of the rod increases.

Therefore, the force required to extend the length of a rod by a specific

amount increases as the cross sectional area of the rod increases,

indicating that the rod with the largest cross sectional area require the

most force and therefore, experiences the largest force.

The rod that experiences the largest force is the rod with the largest cross

sectional area, which is rod B

Learn more here:

brainly.com/question/12937199

4 0

2 years ago

A golf ball has a mass of 46 grams, and when struck by a driver reaches a speed of 42 m/s in a time of 0.00050 seconds. what was

Fittoniya [83]

Impulse in physics is the integral of force, F, with respect with time, t. This value is a vector quantity since force is a vector quantity as well. It can be calculated from the product of force and time. We do as follows:

Impulse = Ft
= m(a)(t)
= m(v/t)t
= 0.046 (42/0.0005) (0.0005)
= 1.932 N-s

4 0

3 years ago

Sound, light, feelings and ideas are not

taurus [48]

I think it’s A, I’m so sorry if I’m wrong.

6 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[Assume that this experiment takes place in deep space so that the effect of gravity is negligible.]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago