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3 years ago

Why does the sky change colors at sunset?

Physics

1 answer:

padilas [110]3 years ago

8 0

The sky changes colors at sunset because the atmosphere ABSORBS some colors of light more than other colors.

The second choice ("The atmosphere bends light") is a correct statement, but it's not the reason that the sky changes colors at sunset.

You might be interested in

Which statement correctly differentiates between transmitters and receivers?

jeka94

Answer:

Transmitters send radio waves, and receivers capture radio waves.

Explanation:

Let us look at each of the choices one by one:

(1).Transmitters have antennas, and receivers do not have antennas.

Nope. To send signals transmitters need antennas, and to receive signals the receivers need antennas as well.

(2). Transmitters send radio waves, and receivers capture radio waves.

This is true. Transmitters are for transmitting and receivers are for receiving EM signals.

(3). Transmitters have demodulators, and receivers have modulators.

No, it is the other way around. Transmitters have modulators, and receivers have demodulators.

(4). Transmitters do not have amplifiers, and receivers have amplifiers.

Nope. Both the transmitters and the receivers need amplifiers. Transmitters need them to increase the power of the broadcast, and receivers need them to amplify the signal for processing.

Therefore, only the 2nd statement "Transmitters send radio waves, and receivers capture radio waves." is correct.

7 0

2 years ago

Read 2 more answers

Do you know any good books about physics and math?

horrorfan [7]

What Is Mathematics? Is a book about math

A book about physics /A Brief History of Time

5 0

2 years ago

A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the

sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0

2 years ago

Read 2 more answers

An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction

Allisa [31]

In this collision, 1/2 of the initial kinetic energy of the first glider is converted into thermal energy.

<h3>In plain English, what is kinetic energy?</h3>

An object's strength as a result ofstrength an object has as a result or motion is known as kinetic energy. Toorder to accelerate an object, a force must be applied. Applying force requires effort on our part. When the work is done, power is transported to the thing, which causes it to move at the athe new, constant pace.

<h3>What does kinetic energy mean, or what are some instances?</h3>

The motion energy is known as kinetic energy, and it is manifested when a particle, object, or group if particles moves. Any moving object uses kinetic energy, including people walking, baseballs being thrown, food falling from tables, and charged particles in electric fields.

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8 0

1 year ago

I've got an energy and work problem. The premise of the problem is:

Alenkasestr [34]

Refer to the diagram shown below.

μ = the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)

6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)

7 0

3 years ago