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3 years ago

What information do you think the temperatures of stars give us?

Physics

2 answers:

kifflom [539]3 years ago

8 0

Answer:

Explanation:

One way of classifying stars is by their temperature .

Science strives to be able to describe how stars and planets form and evolve. This requires theories to describe the processes which include:

Star and planet formation

Star and planet composition

Stellar and solar system evolution

The nuclear processes happening inside stars

The scientific method means that all theories are put to the test. By measuring or calculating the temperature, age and composition of other planets and stars the theories can be tested. If observed values of these parameters are not predicted by theories, then the theories are wrong and need to be revised or replaced.

Read more on Brainly.com - brainly.com/question/16112520#readmore

Paul [167]3 years ago

6 0

Explanation:

One way of classifying stars is by their temperature .

Science strives to be able to describe how stars and planets form and evolve. This requires theories to describe the processes which include:

Star and planet formation

Star and planet composition

Stellar and solar system evolution

The nuclear processes happening inside stars

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Since the Sun has more mass, why do objects on earth not move closer to the Sun instead of staying put on Earth?

maw [93]

Answer:

Because the Earth has it's own gravity that keeps us put, and we also have the moon.

Explanation:

6 0

3 years ago

You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro

guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0

4 years ago

Read 2 more answers

a flashlight has a resistance of 2.4 ohms. what voltage is applied by the batteries if the current in the circuit is 2.5A?

Mashcka [7]

E = I R

That means

Voltage = (current) x (resistance)

= (2.5 A) x (2.4 ohms)

= 6 volts .

8 0

4 years ago

Read 2 more answers

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$