All categories

3 years ago

A blackbody curve relates the wavelength of emitted light to _____. intensity color frequency speed

Physics

2 answers:

valina [46]3 years ago

7 0

Answer: A

Explanation: I got it right on my test

Brainliest?

Kitty [74]3 years ago

4 0

The best answer would be intensity. A black body curve relates to the wavelength of the emitted light to its intensity. To expands the thought, the curve is referring to the power that is being transferred per unit area.

You might be interested in

Plzzzzzzzzzz!!!!!!! Hurryyyyy

Scilla [17]

Answer:

student A or B

Explanation:

A common demonstration is to put a ringing alarm clock or bell in the bell jar, and when the vacuum is created, you can no longer hear the sound of the clock/bell.

The bell is connected to a lab pack or batteries and rung to show pupils it can be heard under normal circumstances. The bell jar is then connected to a vacuum pump using a vacuum plate (see Fig 2) and the air is removed from inside creating a near vacuum. The bell is then again rung. This time however, it cannot be heard.

Small low voltage buzzers can be used as a bell replacement for the bell and work in exactly the same way though teachers generally prefer bells as students may be able to see the hammer moving, proving that it is actually ringing even though they cannot hear it.

Some vacuum pumps are better than others at keeping a strong vacuum though if you cannot completely lose the sound, you will at least notice the volume decreasing.

Sound is simply a series of longitudinal waves travelling from the source, through the air to our ears. Without air present, these waves cannot form and therefore sound cannot be conveyed.

In a longitudinal wave the particles oscillate back and forth in the direction of the wave movement unlike transverse waves which like waves on the sea, single particles travel up and down and not in the direction of the wave.

Because you will not be able to create a perfect vacuum, you may still be able to hear the bell ring slightly. Vibrations from the ringing bell can also travel up to the bung in the bell jar which in turn may resonate the jar slightly. This means you may hear the bell ring, however strong the vacuum. To compensate for this, try to insulate the bell as much as possible from the bell jar. Hanging the bell using elastic cord means some of the vibrations will be absorbed by the cord and not be transferred to the bell jar.

3 0

3 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

4 years ago

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp

Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_{i}$ = 60 mph = 26.8224 m/s

Final velocity $V_{f}$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $\frac{1}{2}$ m( $V_{i}$ ² - $V_{f}$ ² )

we substitute

Δk = $\frac{1}{2}$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $\frac{1}{2}$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0

3 years ago

How do i find the weight of the suitcase<br>pls help asap

Citrus2011 [14]

F=W=mg
12N(given)= m*9.8
m= 1.22 kg

3 0

3 years ago

A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit

barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

4 0

4 years ago