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lina2011 [118]
3 years ago
12

How fast would the car need to go to double its kinetic energy? 12 m/s?

Physics
1 answer:
S_A_V [24]3 years ago
7 0
The car is traveling at v=12m/s. Kinetic energy is given by this formula:
E_k=\frac{mv^2}{2}
We can see that if the kinetic energy depends on the speed quadratically.  This means that if you want to increase the kinetic energy x times you would have to increase speed \sqrt{x} times. We conclude from this that car would have to go:
v'=\sqrt2v=\sqrt2\cdot 12=16.97\frac{m}{s}

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Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

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