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3 years ago

Can aurora be found in other planets?

Physics

2 answers:

Veronika [31]3 years ago

8 0

Answer:

Yes

Explanation:

Any planet with a sufficiently dense atmosphere that lies in the path of the solar wind will have aurora. It has been discoveres in planets like Jupiter, Saturn, etc.

spin [16.1K]3 years ago

4 0

Answer:

yes

Explanation:

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Vygotsky’s theory of cognitive development states that an individual constructs knowledge primarily through interaction with imp

defon

Answer:

false

Explanation:

trust me i j took the test

4 0

3 years ago

Read 2 more answers

In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init

liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,

n = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,

U = − 0.061 J

The charge on an electron,

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:

q ₁ = n q

= 1 ×10 ¹³ * − 1.6 × 10 ⁻¹⁹

= -1.6 × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6 × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and q ₂ are the two charges.

r is the distance between the charge and the point.

K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 = (8.98755 × 10 ⁹ * (-1.6 × 10⁻⁶)²) / r

r = (18.41 × 10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0

3 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago

Can you answer my questions. Please help me

earnstyle [38]

Answer:

answered it.................

8 0

3 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

$a=4.2 m/s^2$

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

$v=0+(4.2)(6.5)=27.3 m/s$

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

$d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m$