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jekas [21]
3 years ago
6

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0mph in time 1.50s .
Part A

At full power, how long would it take for the car to accelerate from 0 to 58.0mph ? Neglect friction and air resistance. THE ANSWER IS "6 seconds" according the mastering physics.. BUT WHY???? Please show how you derive that answer...

Part B

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

Express your answer numerically, in seconds THE ANSWER IS 3 seconds. Please show how you derrive this answer!

Express your answer in seconds.
Physics
1 answer:
kirill115 [55]3 years ago
6 0

Answer:

Part A

it would take 6 sec

it would take 3 sec

Explanation:

We are told that the power supplied to the wheel is constant which means that the sport car is gaining energy i.e

                           power \ \alpha  \ energy

Hence if power is constantly supplied energy constantly increase

From the formula of the Kinetic energy

                       KE  = \frac{1}{2} mv^2

we can see that as the speed doubles from 29 mph  to 58 mph  the energy needed is 2^2 = 4 times of the energy from the formula

   Also the time needed would also be 4 times because energy i directly proportional to time

       Hence to reach 58mph the time that it would take is

                          = 4* 1.5sec = 6sec

     

We are told that the ground pushes the car  with a constant force and

                 F = ma

this means that the acceleration is also constant

             now from newtons law

     v = u +at  

 Looking at it we see that final velocity is directly proportional with time

hence it would take twice the time to reach twice the final velocity

        Time to reach 58mph = 3 s

        since time to reach 29 mph(\frac{1}{2} \ of \ [58mph]) =( \frac{1}{2} \ of \ 3sec )1.5 s

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<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

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  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
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Now, putting the above values

Q_A=Q_B

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

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