Suppose a moving car has 750 J of KE. If the car's speed doubles, how much KE will it have?

If the speed is doubled then the kinetic energy will be 3000J.

Explanation:

The kinetic energy is calculated by ,

KE =(1/2) (M x V x V)

750 = (1/2) (M x V x V)

750(2) = M x V x V

1500 = M x V x V

If the speed is doubled then the kinetic energy will be

KE = (1/2) ( M x 2V x 2V)

KE = (1/2) (4) (M x V x V)

The value of (M x V x V) = 1500

So,

KE = (1/2) (4) (1500)

KE = 3000J

If the speed is doubled then the Kinetic energy will be 3000J

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y