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Readme [11.4K]
3 years ago
11

You start driving your car when the air temperature is 270.734 K. The air pressure in the tires is 454.518 kPa. After driving a

while, the tire's air pressure increases to 470.361 kPa. What is the temperature of the tires at that point, assuming the volume remains constant
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

T₂ =  280.17 K

Explanation:

Here, we can use the equation of state to find the final temperature of air in tires. Equation of State is written as follows:

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Since, the volume is constant.

Therefore,

V_{1} = V_{2} = V\\\\\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\T_{2} = \frac{T_{1}P_{2}}{P_{1}}

where,

T₂ = Final Temperature of Air in Tire= ?

T₁ = Initial Temperature of Air = 270.734 K

P₁ = Initial Pressure of Air = 454.518 KPa

P₂ = Final Pressure of Air = 470.361 KPa

Therefore,

T_{2} = \frac{(270.734\ K)(470.361\ KPa)}{454.518\ KPa}

<u>T₂ =  280.17 K</u>

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4 years ago
Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m
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4 0
3 years ago
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Answer:

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b) \alpha =-35º

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c) The change in the total kinetic energy is:

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ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
3 years ago
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