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jok3333 [9.3K]
3 years ago
14

A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies

, even if the stellar radius and distance are unknown.
b) Explain why in practice this method does not work well if both frequencies are on theRayleigh-Jeans side of the spectrum,hνkT.
c) Derive a simple, approximate expression for the temperature when both measurements areon the Wien tail,hνkT.
d) Derive an expression for the star’s radius if a distance measurement is also available (e.g.,from parallax).

Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

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______is the rate of change in velocity.
shepuryov [24]

Answer:

Acceleration

Explanation:

Acceleration is the rate of change in velocity

speed is how fast an object is moving

Velocity is how fast an object is moving in the particular direction

pls mark me brainliest

6 0
3 years ago
Which one of the following statements is false?
emmasim [6.3K]

Answer:

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Explanation:

Affirmations

a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics

b) True. If both give the same results and use the same quantum number (n)

c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

7 0
3 years ago
1) A force of 157 N is applied to a box 25o above the horizontal. If it is applied over a distance of 14.5 m how much work is do
vfiekz [6]

#1

Work done is given by

W = F.d cos\theta

here we know that

F = 157 N

d = 14.5

\theta = 25^0

now from the above formula

W = (157)(14.5)cos25 = 2063.2 J

#2

By energy conservation

Initial total potential energy of egg = final total kinetic energy of egg

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.8)(8.2)}

v = 12.7 m/s

#3

By momentum conservation we know that

m_1v_1 = (m_1 + m_2)v

215(9.5) = (215 + 19.7) v

v = \frac{215 (9.5)}{215 + 19.7}

v = 8.7 m/s

#4

by momentum conservation we know that

m_1v_1 + m_2v_2 = (m_1 + m_2) v

let direction of velocity of Jay Z car motion is positive direction

now we will say

1100(-20) + 1475v = (1100 + 1475)(7)

1475 v = 2575(7) + 1100(20)

v = 27.14 m/s

Yes Jay Z is overspeeding as his speed is more than speed limit

#5

Tension force in the string will be centripetal force

so here we can say

T = \frac{mv^2}{R}

here we know that

m = 0.63 kg

v = 7.8 m/s

R = 0.23 m

T = \frac{0.63 (7.8)^2}{0.23}

T = 166.6 N

6 0
3 years ago
A dog can hear sounds in the range from 15 to 50,000 Hz. What wavelength corresponds to the upper cut-off point of the sounds at
Svet_ta [14]

Give that,

The frequency range the dog can hear is 15Hz to 50,000Hz

The wavelength of sound in air at 20°C =?

Speed of sound is 344

The frequency corresponding to the lower cut-off point is the lowest frequency which his 15Hz

F=15Hz

The relationship between the wavelength, speed and frequency is given as

v=fλ

Then,

λ=v/f

λ=v/f

λ=344/15

λ=22.93m

6 0
2 years ago
Read 2 more answers
if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet
klio [65]

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

Where:

Q  is the Volume flow rate

r=23 cm \frac{1 m}{100 cm}=0.23 m  is the effective radius

L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

\Delta P=1.88(10)^{4} Pa  is the difference in pressure

\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

4 0
3 years ago
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