Spinning frisbee

By formula of K. E. ,K. E is directly proportional to square of velocity since velocity is doubled As given Kinetic energy is quadruple

The formula for kinetic energy is $KE = \frac{1}{2} \cdot m \cdot v^2$ , where $m$ represents the mass and $v$ represents the the velocity of the object (in this case, the frisbee)

The frisbee's mass stays constant (it doesn't get heavier or lighter while it is being spun), while the velocity doubles when it is being spun at the side, as given in the problem.

Since the velocity doubles and the rest of the terms stay the same, the factor of the kinetic energy is increased by $v^2=2^2=\boxed{4}$ .

Because Velocity have square so if the ratio is 2 so, increase 2^2 = 4

Since $E = \dfrac12mv^2,$ increasing the velocity by a factor of $2$ increases the kinetic energy by a factor of $2^2 = \boxed{4}.$

we have

E = (1/2)mv2

now if 'v' is doubled the kinetic energy will be

E' = (1/2)m.(2v)2

= 4.[(1/2)mv2]

so,

kinetic energy will become four times.

4 times

k.e1 = 1/2mv^2; frisbee spins twice as fast, therefore, velocity increases by a factor of 2; k.e2 = 1/2m(2v)^2 = 4 x 1/2mv^2; hence kinetic energy increases by a factor of 4

it means kinetic energy increases 4 times

since K.E=(0.5) m v^2 and when speed become double of original speed then v=2v so, K.E=(0.5) m(2v)^2=(0.5) m 4v^2 =4 K.E

.

we have 1/2mv^2 we've been given v=2times therefore4[1/2mv^2]

we have E = (1/2)mv2 now if 'v' is doubled the kinetic energy will be E' = (1/2)m.(2v)2 = 4.[(1/2)mv2] so, kinetic energy will become four times.