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2 years ago

Compare the events in the life of the Sun with those of a star that starts with less

Physics

1 answer:

GarryVolchara [31]2 years ago

6 0

A star with greater mass will die out faster than the Sun.

<h3>What factors star is dependent on?</h3>

A star's future relies upon its mass. For the most part, the more huge the star, the quicker it consumes its fuel supply, and the more limited its life. The most huge stars can wear out and detonate in a cosmic explosion after two or three million years of combination.

Our Sun is a typical estimated star: there are more modest stars and bigger stars, even up to multiple times bigger. Numerous other planetary groups have different suns, while our own simply has one. The Sun is made for the most part out of hydrogen and helium gas.

In this manner, one correlation in the occasions in the existence of the Sun with those of a star that beginnings with a mass multiple times more prominent than the Sun's is a star that has a more noteworthy mass will vanish quicker.

Learn more about Star.

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3 years ago

Read 2 more answers

Calculate the number of moles in 100g of water

dsp73

5.55 mol H2O

Explanation:

Water has a molar mass of 18.01528 g/mol. We can then calculate the number of moles of water as

100 g H20 × (1 mol H2O/18.01528 g H20)

= 5.55 mol H2O

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2 years ago

Calculate the pH of hydronium ions in a 0.000 01 M solution of the strong acid HI.

enyata [817]

<h3>Answer:</h3>

pH = 5

<h3>Explanation:</h3>

<u>We are given;</u>

Concentration of a solution HI as 0.00001 M

We are required to calculate the pH of Hydronium ions.

When the acid dissociates in water to;

HI + H₂O → H₃O⁺(aq) + I⁻(aq)

Therefore;

The concentration of H₃O⁺ ions is 0.00001 M\

We need to know that pH = -log[H₃O⁺]

Therefore;

pH =-log 0.00001 M

= 5

Thus, the pH of the hydronium ions is 5

7 0

2 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

3 years ago