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Vika [28.1K]
3 years ago
9

A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw

een the tires and the horizontal road is 0.735, how long will the skid marks be
Physics
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

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Answer:

The value is  P_2 = 40.54 \ psla

Explanation:

From the question we are told that

 The initial pressure is  P_1 = 14\  psla

  The initial temperature is  T_1 =  50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283  \  K

   The final temperature is  T_2 =  320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433  \  K

Generally the equation for adiabatic process is mathematically represented as

         PT^{\frac{\gamma}{1- \gamma} } =  Constant

=>      P_1T_1^{\frac{\gamma}{1- \gamma} } =  P_2T_2^{\frac{\gamma}{1- \gamma} }

Generally for a monoatomic gas  \gamma =  \frac{5}{3}

So

           14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }

=>       14 * 283^{-2.5} =P_2 * 433^{-2.5}

=>       P_2 = 40.54 \ psla

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