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4 years ago

What Type of relationship exists between the Temperature of a Star and the Wavelength of a Star?

2 answers:

8 0

To estimate the surface temperature of a star, we can use the known relationship between the temperature of a blackbody, and the wavelength of light were its spectrum peaks. That is, as you increase the temperature of a blackbody, the peak of its spectrum moves to shorter (bluer) wavelengths of light.

Debora [2.8K]4 years ago

3 0

This is known as Wien's Law:
The relationship is:
wavelength = 0.0029/temperature

It is an inversely proportional relationship.

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Which was not a feature of the northwest ordinance of 1787

Tamiku [17]

What are the choices?

4 0

4 years ago

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

3 years ago

A student conduct an experiment to determine habitation of salt to water affect the density of the water the student feels three

Nikolay [14]

Any like answer choices would be helpful '-'

8 0

3 years ago

Use the claim to answer the question.

Pachacha [2.7K]

Scientific evidence such as reflection and photoelectric effect has proven the wave-particle model of electromagnetic radiation to be true.

<h3>What is electromagnet radiation?</h3>

Electromagnetic radiation is radiation produced as a result of the interactions of the electric and magnetic fields.

Some forms of electromagnetic radiation include:

radio waves
microwaves
ultraviolet radiation
visible light

Electromagnetic radiation can be described using a wave model or a particle model.

The wave property of electromagnetic radiation include:

diffraction
reflection
refraction

The particle property of electromagnetic radiation include:

photoelectric effect
Compton effect.

Therefore, the wave-particle model of electromagnetic radiation is correct because it can be backed up by experiments and evidence.

Learn more about wave-particle model at: brainly.com/question/20452331

6 0

2 years ago

Read 2 more answers

A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6

Natasha2012 [34]

Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

4 0

3 years ago