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Formula for velocity
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
Formula for GPE. Formula for KE.
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
An object dropped from __ would have the most GPE compared to the others.
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
Rate of Acceleration due to Gravity
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
If a roller coaster has 700J of PE at the top of the first hill, how much energy will it have at the bottom of the hill?
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
If you want a car to roll down a hill, which would spot would create the most GPE in the car? Placing it
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
Two trucks with the same mass collide. One truck was moving at 75mph and the other at 60mph. Which had the greatest KE upon impact
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
As a ball falls from the sky after being hit by a bat the energy conversion is from ___
v=d/t
GPE to KE
9.8m/s/s
700J KE
at the peak of the hill
truck traveling 75mph
40m
mgh .5mv^2
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