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Effectus [21]
3 years ago
9

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

increases its speed at uniform rate of at ≡ d |v| dt = 5.22 m/s 2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 25.6 m/s, what is the coefficient of static friction between the tires and the road?
Physics
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) =  mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

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Explanation:

Capacitance of capacitor is given by the expression

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3 years ago
How long does it take a 22 kW steam engine to do 5.6 × 107 J of work? Answer in units of s.
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<h2>Time needed is 2545.45 seconds.</h2>

Explanation:

We know equation for power

               \texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}

Here we need to find time when for a 22 kW steam engine to do 5.6 × 10⁷ J of work.

So

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Substituting

        \texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}\\\\22\times 10^3=\frac{5.6\times 10^7}{\texttt{Time}}\\\\\texttt{Time}=2545.45s

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Answer:

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Explanation:

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Similarly, when he move back to home, time taken is equal to \dfrac{d}{y}

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d=\dfrac{t}{(\dfrac{1}{x}+\dfrac{1}{y})}

d=\dfrac{xyt}{(x+y)}

So, the distance he speed in walking and jogging is \dfrac{xyt}{(x+y)}. Hence, this is the required solution.

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