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3 years ago

At which life cycle stage are stars most stable

Physics

2 answers:

malfutka [58]3 years ago

8 0

The answer is the Young star stage. Hope this helps!

asambeis [7]3 years ago

4 0

~Hello there!

Your question: At which life cycle stage are stars most stable?

Your answer: The young star stage is the life cycle where the stars are most stable because the stars are just flaming, no interference nor there is no disturbance.

Any queries ^?

Happy Studying!

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A metal container with the oil weigh

MA_775_DIABLO [31]

Answer:

16 kg

Explanation:

M - container

m - oil mass

by definition of density ,

relative density is the ratio of the density of a substance to the density of water.

relative density = density/ density of water

density of oil = 1.2*1000 kgm⁻³ = 1200 kgm⁻³

1 Litre =10⁻³ m³

oil volume = 80 *10⁻³ m³

mass of oil = density * volume

= 1200*80*10⁻³

= 96 kg

Mass of container + mass of oil =112

mass of container = 112 - 96

= 16 kg

7 0

3 years ago

What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af

sladkih [1.3K]

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$ .

4 0

2 years ago

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

galben [10]

3 meters per second

5 0

3 years ago

If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

$Q=ne$

e is the charge on electron

$n=\dfrac{Q}{e}$

$n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7.5625\times 10^{11}$

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0

3 years ago

What is the trend noticed with the inner planets

lutik1710 [3]

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7 0

3 years ago