What is the change in kinetic energy of a baseball thrown at 90 mph compared to one thrown at 45 mph?
it is four times as much right?
Yes, four times is correct.
Kinetic energy is calculated according to the formula:
KE = (1/2)mv²
So if v2/v1=90/45=2
KE2/KE1
= (1/2)mv2²/((1/2)mv1²)
= (v2/v1)²
= 4
In fact, the "change" is
KE2-KE1
=(1/2)m(v2²-v1²)
=(1/2)m((2*v1)²-v1²)
=3*(1/2)mv1²
I.e. the "change" in KE is three times that of the slower ball.
In fact, I would use the word "difference" and not "change", since the balls are distinct.
ok, thanks for your help
To determine the change in kinetic energy, we need to calculate the kinetic energy for each baseball and then compare the two values.
The formula for kinetic energy is:
Kinetic Energy = (1/2) * mass * velocity^2
Let's assume that the mass of both baseballs is the same.
For the baseball thrown at 90 mph:
Kinetic Energy1 = (1/2) * mass * (90 mph)^2
And for the baseball thrown at 45 mph:
Kinetic Energy2 = (1/2) * mass * (45 mph)^2
To compare the two values, we can calculate the ratio of their kinetic energies:
Ratio = Kinetic Energy1 / Kinetic Energy2
If the ratio is 4, then the change in kinetic energy for the baseball thrown at 90 mph compared to the one thrown at 45 mph is indeed four times as much.
However, in order to confirm this, we would need to calculate the actual values of Kinetic Energy1 and Kinetic Energy2, not just the ratio. This would require knowing the mass of the baseballs, as well as converting the velocities from mph to a more appropriate unit, such as meters per second (m/s).