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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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What is meant by resistance and voltmeter?

LUCKY_DIMON [66]

Answer:

The internal resistance of an ideal ammeter will be zero since it should allow current to pass through it. Voltmeter measures the potential difference, it is connected in parallel. .

Explanation:

<h3>I hope this helps!</h3>

7 0

1 year ago

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

5 0

2 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

What is kinetic energy

mixas84 [53]

(symbol K) Energy that an object possesses because it is in motion. It is the energy given to an object to set it in motion; it depends on the mass (m) of the object and its velocity (v), according to the equation K = 1/2 mv2. On impact, it is converted into other forms of energy such as heat, sound and light.

8 0

2 years ago

Question:

KengaRu [80]

Answers:

1A) Al203

1B) SF6

2) Fe203 - iron oxide

3 0

3 years ago