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Andre45 [30]
3 years ago
10

As a clump of interstellar gas contracts to become a main-sequence star, its changing position on the H-R diagram tells us _____

_____.
Physics
1 answer:
Stels [109]3 years ago
6 0

Answer:

how its outward appearance is changing

Explanation:

H-R diagram is a fundamental tool used to classify stars. The position of stars on the H-R diagram is dependent on its luminosity and temperature.  

The position of the star is represented by its outward appearance specially the color. The hottest and brightest the star, the more it lies between the red and blue color range.  

The cooler stars lie on the right side of the H-R diagram while the hot stars lie on the left side of the H-R diagram.  

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A 2.24x10^3 kg car slows down uniformly from 20.0 m/s to a stop (assume constant acceleration) via a braking force of 8.41x10^3
melisa1 [442]

Answer:

Explanation: 139 m

4 0
3 years ago
Is anyone good at science I need help with 2 tests
Katyanochek1 [597]

Answer:

i am!

Explanation:

7 0
3 years ago
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad
Anna71 [15]

Answer:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

a_{r} = \omega^{2}\cdot R

Where:

\omega - Angular speed, measured in radians per second.

R - Radius of rotation, measured in meters.

The angular speed is first determined:

\omega = \frac{\pi}{30}\cdot \dot n

Where \dot n is the angular speed, measured in revolutions per minute.

If \dot n = 3000\,rpm, the angular speed measured in radians per second is:

\omega = \frac{\pi}{30}\cdot (3000\,rpm)

\omega \approx 314.159\,\frac{rad}{s}

Now, if \omega = 314.159\,\frac{rad}{s} and R = 0.1\,m, the resultant acceleration is then:

a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)

a_{r} = 9869.588\,\frac{m}{s^{2}}

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

6 0
3 years ago
9. [03.03]
Evgen [1.6K]

Answer:

Circuit one will have more current than circuit two

Explanation:

I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm's Law, which states the following -

V = IR,

where V = voltage / potential difference, I = current, and R = resistance

If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance -

2V = I( 1 / 2R ),

4V = IR,

I = 4V / R

Whereas in the second circuit -

V = IR,

I = V / R

As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

<u><em>Hence, circuit one will have more current than circuit two</em></u>

5 0
4 years ago
Read 2 more answers
If you know how much radioactive material the organism had to begin with, explain how you could use half-life to determine its a
TiliK225 [7]

Answer: We can calculate it with the radioactive half life equation

Explanation:

If we already know the initial amount of radioactive material and its half life, we can leave that material for a specific known time and then measure how much of the material is left (since it follows the radioactive deacay) and use the results in the following formula:

A=A_{o}.2^{\frac{-t}{h}}  

Where:

A is the final amount of the material

A_{o} is the initial amount of the material

t is the time elapsed

h is the half life of the radioactive compound

5 0
3 years ago
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