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inn [45]
4 years ago
15

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

Physics
2 answers:
Nesterboy [21]4 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]4 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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A student throws a pumpkin up from 14.2 meters with a speed of 19.4m/s. what is the velocity
RSB [31]

.73 because velocity is determined by distance divide by speed brainliest if this helps


6 0
4 years ago
What is the main reason why many nuclear power plants are located near bodies of water? to wash wastes out of the power plant to
Natalija [7]

Answer: to avoid problems with water supply

Explanation: power plant needs water to run

7 0
3 years ago
Read 2 more answers
Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

5 0
3 years ago
True or false? An apple, potato, and onion all taste the same if you eat them with your nose plugged
aleksandr82 [10.1K]

Answer:

True

Explanation:

3 0
3 years ago
Read 2 more answers
An external resistor with resistance R is connected to a battery that has an emf E and an internal resistance r. Let P be the el
satela [25.4K]

Answer:

a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

Explanation:

<u>Solution  :</u>

(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation  in the next form  

P=∈*I-I^2*r                (1)

Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by  

I=∈/R+r                     (2)

When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

I= ∈/r

Now let us plug this expression of I into equation (1) to get the consumed power  

P=∈*I-I^2*r

 =I(∈-I*r)

 =0

The consumed power when R is very small is zero  

(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes  

I=∈/R

The dissipated power due toll could be calculated by using equation.

P=I^2*r                (3)

Now let us plug the expression of I into equation (3) to get P  

P=I^2*R=(∈/R)^2*R

 =∈^2/R

4 0
3 years ago
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