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5 questions
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  • Q1
    What variables are given in the situation above?
    Question Image
    PE=KE=15,000J, m=0.5kg, x=0.10m
    PE=KE=0.5J, m=0.10kg, x=15,000m
    PE=KE=0.5J, m=15,000kg, x=0.10m
    PE=KE=15,000J, m=0.10kg, x=0.5m
    120s
    112.39.c.6.a
  • Q2
    What equation can be used to solve for the spring constant?
    Question Image
    W=FD
    PE=(1/2)kx^2
    KE=(1/2)mv^2
    PE=mgh
    120s
    112.39.c.6.a
  • Q3
    What is the value of the spring constant for the tennis ball launcher according to the data provided?
    Question Image
    3,000,000N/m
    1,732N/m
    3.6x10^13N/m
    6,000,000N/m
    120s
    112.39.c.6.a
  • Q4
    If the mass of the ball was doubled, how would the spring constant change?
    Question Image
    No change
    Quadruple
    Halve
    Double
    120s
    112.39.c.6.a
  • Q5
    What would a graph of Potential energy vs distance compressed look like?
    Question Image
    Straight line with positive slope
    Straight line with negative slope
    Curved line with positive increasing slope
    Curved line with positive decreasing slope
    120s
    112.39.c.6.a

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