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Nina [5.8K]
3 years ago
5

Which best describes how energy transferred from an electron in the solar wind compares with energy absorbed by an electron in t

he oxygen atom?
equal to

less than

No energy is transferred.

greater than
Physics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

equal to

(sorry if this is late)

Explanation:

Rest of the answers.

As solar wind approaches Earth, what happens to the charged electrons?

They are deflected as Earth’s magnetic field exerts force on them.

Which best describes how energy transferred from an electron in the solar wind compares with energy absorbed by an electron in the oxygen atom?

equal to

Which statement is accurate about how the aurora borealis is formed?

When the electrons fall to a lower energy state, they release energy as electromagnetic radiation, light.

How does the energy in the light of the aurora borealis compare to the energy as an excited electron returns to its original energy level?

equal to

During the formation of the aurora borealis, the electrons in an atom experience a change in energy levels. Which statement about this change is accurate?

First, the electron absorbs energy to move to a higher energy level.

You might be interested in
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
Engineers tasked with building a car bumper need high-quality plastic that is readily available. They found the material polycar
jenyasd209 [6]

Answer:

The material must be durable (quality of the material requirement)

Explanation:

The design criteria set for the materials used for technological design are;

1) The materials should be affordable (less costly)

2) The materials should be last for a long duration (high durability)

3) The material should be readily available (easily sourced)

Therefore, given that the engineers initially had the criteria for the required plastic to be of high quality and to be readily available, and that the poly-carbonate they found is long lasting and not too costly, the criteria met that was set initially was the  quality criteria of durability.

7 0
3 years ago
Read 2 more answers
What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation
Lorico [155]
The plant will not grow. In fact it could have all the nutrients and all the water it needs, but without a sufficient amount of light, it could die because its leaves are meant for a certain minimum amount of light.

I'll come back and see if you have posted the question you wanted and edit my answer.
6 0
3 years ago
You have a motion security light by your door cat runs across the in the light turns on what is the input of the system and what
Neporo4naja [7]
A motion security line is a system that is used to detect motion.
The input for the system is MOTION and the output is LIGHT. That is, when the system detects motion it switch on light.
Remember, an Input is the information that was inserted into a system while the output is the result of the processed information.
5 0
3 years ago
An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
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