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3 years ago

An aurora occurs when ____

1 answer:

7 0

The short answer to how the aurora happens is that energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.

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What is potential energy called that is associated with objects that can be stretched

Umnica [9.8K]

Elastic potential energy

5 0

3 years ago

Savion listed the steps involved when nuclear power plants generate electricity.

OverLord2011 [107]

Answer:

Reorder the steps so that step 4 appears before step 3

Explanation:

In a nuclear power plant, we have;

1) Nuclear reaction between the radio active species and the particles takes place to generate energy in the nucleus of atoms

2) The nuclear energy in the atom is converted into radiant energy, which is the energy found in light, and thermal (heat) energy

3) The produced radiant and thermal energy is released as heat and light

4) With the produced heat, steam is generated

5) The generated steam turns the steam turbines and produced mechanical energy

6) The produced mechanical energy is then converted into electrical energy in the electrical generator of the power plant

To correct Savion's error, Step 4) the light and heat should be released before step 3) the released heat can be used to generate steam, we therefore reorder the steps so that step 4 appears before step 3.

4 0

3 years ago

Read 2 more answers

Block 1 (mass 2.00 kg) is moving rightward at 10.0 m/s and block 2 (mass 5.00 kg) is moving rightward at 3.00 m/s. The surface i

DaniilM [7]

Answer:

a) 0.25m

b) 5 m/s

Explanation:

When the spring is compressed both boxes are moving with the same velocity, so applying the principle of linear momentum conservation:

$m1*v_{o1}+m2*v_{o2}=(m1+m2)*v\\v=5m/s$

Now applying the principle of energy conservation:

$K1+K2+U_{g1}-U_e=Kf+U_{g2}\\K1+0-U_e=K2+0\\U_e=K1+K2-kf\\\frac{1}{2}*k*x^2+=\frac{1}{2}*m1*v1^2+\frac{1}{2}*m1*v1^2-\frac{1}{2}*(m1+m2)*v^2\\\\x=\sqrt{\frac{2.00kg*(10m/s)^2+5.00kg*(3.00m/s)^2-7.00kg*(5m/s)^2}{1120N/m}}\\x=0.25m$

We got that the maximum compression is 0.25m.

5 0

2 years ago

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one

anygoal [31]

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)

where $g=9.8 m/s^2$ is the acceleration of gravity.

Dividing eq.(1) by eq(2), we get:

$tan \theta = \frac{a}{g}$

And therefore, we find the angle:

$\theta=tan^{-1}(\frac{a}{g})=tan^{-1}(\frac{0.90}{9.8})=5.2^{\circ}$

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is

Charra [1.4K]

Answer:

M is equal to m

Explanation:

In case we say that the green block's mass m is less than red block's mass M, then the green block would have bounced and moved back to the left instead of coming to rest. The other case where if mass of green block's mass m would have been greater than the red block's mass M, the green block would have kept moving to the right instead of coming to rest. After collision, the red block moves to the right because of exchange of velocities. Therefore, m=M since m comes to rest and M moves to the right

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂, which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

8 0

3 years ago