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Tanya [424]
3 years ago
12

A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then trav

els for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Physics
1 answer:
bija089 [108]3 years ago
6 0

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

v_{average}=\frac{155km}{2.25h}=68.9km/h

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The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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