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3 years ago

The temperature of a star is 6000K and its luminosity is 1000. Which type of star is it?

Physics

2 answers:

Nesterboy [21]3 years ago

7 0

Answer:it’s a supergiant

Explanation: if this is what you are doing for your mastery test then this is the correct answer

Annette [7]3 years ago

6 0

Answer:

F-type star

Explanation:

Stars are classified on the basis of temperature and luminosity in the H-R diagram.

Based on the temperature, the stars classified as:

O, B, A, F, G, K, and M.

O-type stars are the hottest stars and M-type stars are the coolest ones.

A star having temperature 6000 K and luminosity 1000 would be a F-type star.

Luminosity depends on temperature and size of a star. More the temperature, more would be its luminosity.

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Having trouble with a physics class problem. Help?

tekilochka [14]

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2 years ago

A 40-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of

ratelena [41]

Answer:

The force the skater exerts is 50 N

Direction: towards wall

Explanation:

The impulse is equal to:

$I=F_{ave} delta-t=m*delta-v=m(v_{f} -v_{i} )$

The average force is:

$F_{ave} =\frac{m*(v_{f}-v_{i}) }{delta-t}$

Where

vf = final speed = -1 m/s

vi = initial speed = 0

m = mass = 40 kg

Δt = time interval = 0.8 s

Replacing:

$F_{ave} =\frac{40*(-1-0)}{0.8} =-50N$ (force applied by the wall)

The force the skater exerts is

Fsk = -Fave = -(-50) = 50 N

Direction: towards wall

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2 years ago

Please help me with this physics question!

Vadim26 [7]

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3 years ago

Who were major innovators of the late nineteenth or early twentieth century?

Nookie1986 [14]

The major innovators in the late nineteenth or early twentieth century isThomas Edison Alexander Graham Bell

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2 years ago

Read 2 more answers

At a sudden contraction in a pipe the diameter changes from D 1 to D 2 . The pressure drop, Δ p , which develops across the cont

Thepotemich [5.8K]

Answer:

Velocity in the smaller pipe should not be included as an additional variable.

Explanation:

$\Delta P = f(D_1, D_2, V, \rho, \mu)$

The dimensional formula of the variables are

$\Delta P = FL^{-2} , D_1 = L, D_2=L, V=LT_{-1}, \rho =FL^{-4}T^2, \mu = FL^{-2}T$

Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.

Use, D_1, V, \mu as the repeating variables.

Therefore, $\pi = \Delta PD_1^aV^b \mu^c$

$(FL^{-2})(L)^a(LT^{-1})^b(FL^{-2}T)^c = F^0L^0T^0$

From this

1+c=0

-2+a+b-2c=0

-b+c=0

c=-1, b = -1, a = 1

Now, $\pi_1=\frac{\Delta PD_1}{V\mu}=\frac{(ML^{-1}T^{-2})L}{(LT^{-1})(ML^{-1}T^{-1})}=M^0L^0T^0$

For $\pi_2 = D_2D_1^aV^b\mu^c$

$F^0L^0T^0=L(L)^a(LT^{-1})^b(FL^{-2}T)^c$

c = 0

1 + a + b - 2c = 0

-b + c = 0

Therefore, a = -1, b = 0, c = 0

$\pi_2 = \frac{D_2}{D_1}$

For $\pi_3$

$\pi_3 = \rho, D_1^a V^b\mu^c$

$F^0L^0T^0 = (FL^{-4}T^2)(L)^a(LT^{-1})^b(FL^{-2}T)^c$

1 + c = 0

-4 + a + b - 2c = 0

2-b+c=0

c=-1, b=1, a = 1

Therefore, $\pi_3 = \frac{\rho D_1V}{\mu}$

Now checking,

$\pi_3 = \frac{(ML^{-3})(L)(LT^{-1})}{ML^{-1}T^{-1}} = M^0L^0T^0$

Therefore, $\frac{\Delta P D_1}{V \mu} = \phi (\frac{D_2}{D_1}, \frac{\rho D_1 V}{\mu})$

From continuity equation

$V\frac{\pi}{4}D_1^2 = V_s \frac{\pi}{4}D_2^2$

$V_s = V (\frac{D_1}{D_2})^2$

$V_s$ is not independent of $D_1,D_2, V$

Therefore it should not be included as an additional variable.

3 0

2 years ago