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oee [108]
3 years ago
5

What is the approximate absolute brightness and temperature of the dwarf star labeled A

Physics
2 answers:
V125BC [204]3 years ago
8 0
Brightness is 10000 and temperature of giant star is 20000
Alik [6]3 years ago
7 0

Answer:

The approximate absolute brightness is 10,000 and the temperature is 22,000 of the main sequence star.

The approximate absolute brightness is 100-1,000 and the temperature of the giant star is 22,000.

The answer is supergiants.

Explanation:

Hope u do good :)

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which measurment describes the amount of heat needed to raise the temperature of one gram of a material by one degree celsius?
ella [17]

<u>Answer:</u>

Specific Heat

<u>Explanation:</u>

Specific heat is the measurement which describes the amount of heat needed to raise the temperature of one gram of a material by one degree Celsius.

It is the amount of heat required per unit mass to raise the temperature by one degree Celsius. The relationship between heat and the temperature change is usually expressed as shown below:

Q=cmΔT

where Q = heat added,

c= specific heat,

m=mass; and

ΔT = change in temperature


7 0
4 years ago
Ignoring air​ resistance, an object in free fall will fall d feet in t​ seconds, where d and t are related by the algebraic mode
Lostsunrise [7]

Explanation:

Given that,

An object in free fall will fall d feet in t​ seconds, where d and t are related by the algebraic model as :

d=16t^2..........(1)

(a) We need to find the time taken by the object to fall 1148 ft. Put this in equation (1) as :

16t^2=1148

t=\sqrt{71.75}

t = 8.47 seconds

(b) If the object is in free fall for 18.5 sec after it is​ dropped, then the height of the object is given by :

d=16(18.5)^2

d = 5476 ft

Hence, this is the required solution.

7 0
4 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
3 years ago
Read 2 more answers
Comparing Wave A (black) to Wave B (green), Wave A has a
Gennadij [26K]
I think it's C, longer wave length.
7 0
4 years ago
Read 2 more answers
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy(KE) when it jumps PLEASE HELP ME
Kay [80]

Answer:

0.91 J

Explanation:

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

For the lemming in this problem, when he jumps, we have:

m = 0.0780 kg is the mass

v = 4.84 m/s is the speed

Substituting into the equation, we find:

K=\frac{1}{2}(0.0780)(4.84)^2=0.91 J

8 0
3 years ago
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