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kodGreya [7K]
3 years ago
10

The auroras occur in the

Physics
1 answer:
kotykmax [81]3 years ago
8 0

Answer:

Ionosphere

Explanation:

The thermosphere reaches 600 kilometres just above mesosphere and begins immediately above the mesosphere. This layer is where the aurora and satellites appear.

The ionosphere is the comprehensive career of the mesosphere because most of the thermosphere, located 80–400 kilometres just above ground atmosphere.

Auroras — magnificent flowing streaks of light seen in the night sky – appear in this location.

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How does sound travel through a medium?
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Particles vibrate parallel to the direction the sound travels. It's a longitudinal wave.
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3 years ago
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Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
6 0
3 years ago
A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0
a_sh-v [17]

Answer:

a. -12.7 Nm

b. -7.9 rad/s^2

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

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F_1 = 90.0N\\F_2 = 125N

Other parameters given include:

Mass of solid disk, M = 24.3kg

and radius of solid disk, r = 0.364m

a.) The formula for determining torque (T), is T = r * F

Hence the net torque produced by the two forces is given as a summation of both forces:

T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm

b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

I_{disk} = {\frac{1}{2}}Mr^2

where M = Mass of solid disk

and r = radius of solid disk

We then relate the torque and angular acceleration (\alpha) with the formula:

T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha  = -{\frac{12.7}{1.61}} = -7.9 rad/s^2

4 0
4 years ago
A rocket is launched at an angle of = 49° above the horizontal with an initial speed vi = 54 m/s, as shown below. It moves for 2
4vir4ik [10]
I never seem to pretty best friends
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3 years ago
A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a
Slav-nsk [51]

For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

6 0
3 years ago
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