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2 years ago

Where is the natural light display called aurora borealis located?

Physics

1 answer:

Xelga [282]2 years ago

5 0

The natural light display called aurora borealis is located in the northern

hemisphere.

There are two types of aurora which are called aurora borealis and aurora

australis. The aurora borealis is located in the Northern hemisphere while

the aurora australis is located in the Southern hemisphere.

They receive their energy through the interaction of charged particles

on the Sun and Earth to produce the light display. An example

of the interaction involves solar wind with atoms of the upper atmosphere.

Read more on brainly.com/question/20191244

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The process of combining two small nuclei into one nucleus of larger mass is called _____.

fiasKO [112]

The process of splitting one large nucleus into
smaller ones is nuclear fission.

The process of combining two small nuclei into
one larger one is nuclear fusion.

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3 years ago

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10) If the mass 2m, the left mass

Romashka-Z-Leto [24]

Answer:

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

Explanation:

Gravitational force between two objects of masses $m_{1}, m_{2}$ kept at a distance r is given by the formula

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$

Here , $m_{1}$ = 2m

$m_{2}$ = $\frac{m}{2}$

Thus , F = $\frac{-G.2m.\frac{m}{2} }{r^{2} }$

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

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3 years ago

Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.

Paraphin [41]

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

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3 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

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3 years ago

Which law states that energy is neither created nor destroyed? a. Law of Conservation of Mass

Brrunno [24]

Answer:

C. Law of Conservation

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3 years ago

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