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yaroslaw [1]
3 years ago
15

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

iles per hour, determine the rate of change of the distance between Hartville and the car when the car has been travelling for 55 miles. (Do not include units in your answer, and round to the nearest hundredth.)

Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

x^2=y^2+145^2

On differentiating

2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}

\dfrac{dx}{dt}=14.89\ miles/hr

Hence, The rate of change of the distance is 14.89.

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Generally solving mathematically

     -(0.45*m* (T_2-100  \textdegree c)) = 0.45*m*(T_2 -55\textdegree c)

     -(T_2-100 \textdegree c)) = (T_2 -55 \textdegree c)

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Explanation:

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Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

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(c)   We know that for an isothermal process,

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

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                    = 28.7 atm

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