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3 years ago

Which layer of the atmosphere is between the mesosphere and the exosphere?

Physics

2 answers:

Llana [10]3 years ago

8 0

Answer:

thermosphere

Explanation:

Alinara [238K]3 years ago

5 0

Answer:

The answer is thermosphere

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Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec

scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0

4 years ago

Un coche de carreras de Fórmula 1 (véase Figura 2.30) acelera desde el reposo a razón de 18 m.s-2 . Suma que se mueve en línea r

Lynna [10]

Answer:

I will answer in English.

Ok, we know that the acceleration is a = 18m/s^2, and we have that the initial velocity and position are both zero. (because it starts at rest)

then we have:

a(t) = 18m/s^2

for the velocity, we integrate over time (because the initial velocity is equal to zero we do not have any integration constant)

v(t) = (18m/s^2)*t

for the position we integrate again over time, and again, we do not have any integration constant

p(t) = (1/2)(18m/s^2)*t^2 = (9m/s^2)*t^2

a) The speed at t= 3s can be found by replacing t = 3s in the velocity equation.

v(3s) = (18m/s^2)*3s = 54m/s

b) the distance traveled by this time can be found by replacing t = 3s in the position equation.

p(3s) =(9m/s^2)*(3s)^2 = 81 m

c) first, we need to find what is the time when the position is equal to 200m.

p(t) = 200m = (9m/s^2)*t^2

√(200/9) s = t = 4.7s

Now we replace that time in the velocity equation and we get:

v (4.7s) = (18m/s^2)*4.7s = 84.6m/s

d) ok, to do this we know that.

1 hour has: 60*60 = 3600 seconds.

then we have the transformation k = 1h/3600s

1 km has 1000 meters.

then we have the transformation c = 1km/1000m

so we have that:

84.6m/s = 84.6m/s*(c/k) = 84.6*(3600/1000)km/h = 304.56 km/h

6 0

3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

melisa1 [442]

Answer:

$H_2 = 91.55 km$

Explanation:

Gravity on the surface of planet is given as

$g = \frac{GM}{R^2}$

as we know that

$M = 8.93 \times 10^{22} kg$

$R = 1821 km$

now gravity on the planet is

$g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}$

so we have

$g = 1.8 m/s^2$

now we know that

$H_{max} = \frac{v^2}{2g}$

so we will say

$\frac{H_1}{H_2} = \frac{g_2}{g_1}$

$\frac{500}{H_2} = \frac{9.81}{1.8}$

$H_2 = 91.55 km$

5 0

3 years ago

the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s

Travka [436]

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

3 0

3 years ago

What would you need to know to calculate the angular momentum of a man on a Ferris wheel?

valentinak56 [21]

The angular momentum calculated with respect to the axis of rotation of an object is given by:
$L=mvr$
where m is the object's mass, v is its tangential speed, and r is its distance from the axis of rotation.

In case of a man on a Ferris wheel, we need to have these quantities in order to calculate his angular momentum. These quantities corresponds to:
- m, the mass of the man
- v, the tangential speed of the wheel at its edge
- r, the radius of the wheel

It is possible to calculate the angular momentum even if we don't know v, the tangential speed. In this case, we need to know at least the angular velocity $\omega$ (because from this relationship we can find the tangential speed: $v=\omega r$ ) or the period of rotation of the wheel, T (because we can find the angular velocity from it: $\omega= \frac{2 \pi}{T}$ ).

4 0

4 years ago