A 2630-kg car drives up a hill 769 meters high in 41.0 seconds. Calculate the power exerted
i am also stumped on this question
The bounce periods y are measured for a set of masses x hung from a spring. Assuming the exponential form y = bx m, (log y) vs. (log x) is plotted, yielding a linear trendline equation of (log y) = 0.4796 (log x) + 0.1029, with R2 = 0.9991. Which of the following exponential functions is most likely?
not enough info for the first one, what angle is the hill at? also, need either a final or initial velocity of some kind
second problem:
what are the choices? it sounds like it is multiple choice
To calculate the power exerted, we'll use the equation:
Power = Work / Time
First, we need to calculate the work done. The work done is equal to the change in potential energy as the car drives up the hill.
Work = mgh
where:
m = mass of the car (2630 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the hill (769 m)
Work = 2630 kg * 9.8 m/s^2 * 769 m
Now, we can substitute the work value and the time value into the power equation:
Power = Work / Time
Power = (2630 kg * 9.8 m/s^2 * 769 m) / 41.0 s
Calculate the numerical value of the expression and round it to the appropriate level of precision to get the final answer.