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alisha [4.7K]
3 years ago
5

Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s

. What temperature does this represent?
Physics
1 answer:
34kurt3 years ago
7 0

Answer:

T = 1.26 \times 10^8 K

Explanation:

As we know that rms speed of ideal gas is given by the formula

v_{rms} = \sqrt{\frac{3RT}{M}}

here we know that

v_{rms} = 2.67 \times 10^5 m/s

molecular mass of gas is given as

M = 44 g/mol = 0.044 kg/mol

now from above formula we have

2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}

now we have

T = 1.26 \times 10^8 K

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If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat
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To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

F = k \frac{q_1q_2}{r^2}

Here,

k = Coulomb's constant

q = Charge of proton and electron

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Replacing we have that,

F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})

F = 3.6956*10^{-9}N

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

a_e = \frac{F}{m_e}

a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}

a_e = 4.0566*10^{21}m/s^2

The acceleration of the proton is given as,

a_p = \frac{F}{m_p}

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ioda

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A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the
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Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

2as = u²

s = u² / 2a

s = (14)² / (2 x 16.333)

s = 6.0 m

Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.

The correct option is d

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