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denis23 [38]
3 years ago
12

Please use mesosphere in a sentence

Physics
2 answers:
RideAnS [48]3 years ago
8 0
The coldest part of the atmosphere is the mesosphere
fiasKO [112]3 years ago
7 0
<span>In the mesosphere, between 50 and 80 km (30 and 50 miles) above the Earth, the temperature drops with altitude, reaching around -100°C (-148°F).</span>
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If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds
s2008m [1.1K]

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(c) The time for the block to hit the surface is 4.03.

(d) The velocity of the block at the maximum height is 0.

<h3>Velocity of the rock</h3>

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

  • g is acceleration due to gravity in mars = 3.72 m/s²

v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

<h3>Velocity of the rock when t = 2 second</h3>

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

<h3>Time for the rock to reach maximum height</h3>

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

<h3>Velocity of the rock when it hits the surface</h3>

v = u - gt

v = 15 - 3.72(4.03)

v = 0

Learn more about velocity at maximum height here: brainly.com/question/14638187

8 0
2 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
1: In a longitudinal wave, the particle displacement is __________ the direction of wave movement
jenyasd209 [6]
The answer is parallel

5 0
3 years ago
Which of the following color schemes (specifically in jewel tones) is on-trend for 2020?
stiks02 [169]

Answer:

The answer is B complimentary

Explanation:

3 0
3 years ago
Read 2 more answers
25 POINTS FOR ANSWER How are Newton’s Laws used to describe the motion of planets? Justify your response in two or more complete
Alexus [3.1K]

Pour la seule et simple raison qu'elle s'exerce entre tous les corps de l'univers ( objet, astres etc..

Si on tient compte des frottements liés aux chocs successifs des billes les une sur les autre, au bout d'un certain temps, le mouvement va cesser.

Si on dit que toute l'énergie potentielle de pesanteur est transformée en énergie cinétique, et réciproquement, donc que l'énergie mécanique est conservée au fil des chocs et des rebonds, alors, le mouvement est perpétuel. Le nombre de billes qui remontent est toujours égal au nombre de billes qu'on a lâchées.

La première loi concerne des systèmes immobiles, ou plutôt on considère des systèmes dit "isolé", c'est à dire qu'ils ne sont pas soumis à d'autre force que celle que l'on connait.

Ce qu'il faut retenir de celui ci c'est ça :

Si j'ai un système en mouvement rectiligne uniforme OU immobile, alors :

Avec F1 F2 F3, trois forces s'exercant sur mon système

Attention ! Ici je n'ai pas mit les flèches sur les différentes forces mais elles sont obligatoires ! On parle de vecteur force !

Pour la deuxième loi :

C'est le même principe, la différence c'est que l'on est en mouvement.

 

Avec a le vecteur accélération. Il y a beaucoup de ressource sur le net, n'hésite pas à regarder, la j'ai simplement pu te donner les expressions les plus connus. Mais il faudra les manipuler, et sans exercice sur lequel se baser, c'est plus difficile ! 

La troisième loi est bien moins importante que les deux autres, mais n'hésite pas à regarder sur le net, tu trouveras l'énoncé. C'est la même logique.

4 0
3 years ago
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