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3 years ago

How are Aurora produced?

Physics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

Answer:

The third one.

Explanation:

Atmospheric molecules releasing photons (light) in response to collisions with Solar Wind particles because when auroras are produced what actually happens is that the ions of the solar wind collide with atoms of oxygen and nitrogen from the Earth's atmosphere. The energy released during these collisions causes a colorful glowing halo around the poles which is the aurora.

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A clindrical rod of uniform density is located with its center at the origin, and its axis along the axis. It rotates about its

zhenek [66]

Answer:

6093.2328 J

Explanation:

For cylindrical rod moment of inertia will be

$I_X=I_Y=\frac{1}{12}m(3r^2+h^2)$

$I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236$ $kg -m^2$

we have given time =0.02 sec

Angular speed = $\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec$

Rotational KE = $\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J$

5 0

3 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

DENIUS [597]

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

$F_g$ = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

$F_d$ = $\frac{1}{2}$ pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

$F_g = F_d$

mg = $\frac{1}{2}$ pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

5 0

3 years ago

Which of these forms of radiation passes most easily through the disk of the Milky Way?

Rama09 [41]

<h2>Answer: 3 - infrared light</h2>

Explanation:

<u>There are certain areas of the Milky Way that cannot be observed using the visible range of the electromagnetic spectrum</u> (this includes blue light and red light). This is because these areas are covered or hidden behind columns of interstellar dust and dark matter.

However, using infrared light and sometimes radio waves, it is possible to observe the galaxy better, because this light manages to pass through all that interstellar dust.

4 0

3 years ago

0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b

andreev551 [17]

Answer:

c. 250k

Explanation:

4 0

3 years ago

4. With respect to a mass of 1kg. (a) How much change in elevation must it undergo to change its potential energy by 1 kJ? (g=9.

mojhsa [17]

Answer:

102 m upwards.

Explanation:

Just from a qualitative analysis we can tell the mass it needs to go upwards. How much we determine with the fact that the increase will be - in absolute value - equal to the work gravity does on it to go down that same distance.

Fixed that work being 1 kJ, we get

$\vec F \cdot \vec{ \Delta h} = |F| |\Delta h| cos 0 = ( 9.8 N )\Delta h \cdot 1 = 10^3 J\\\Delta h = \frac {10^3}{9,8} m \approx 102 m$

4 0

2 years ago