A0.060 kg ball hits the ground with a speed of –32 m/s. the ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 n force on the ball.

what is the magnitude of the velocity after it hits the ground?

9.3 m/s
12 m/s
41 m/s
73 m/s

9.3 m/s

Explanation:

The change in momentum of the ball is equal to the impulse given to the ball, which is equal to the product between the force exerted and the time taken:

where:

F = 55 N is the force exerted on the ball

is the duration of the collision

Substituting, we find the change in momentum of the ball:

The change in momentum of the ball can also be written as:

where:

m = 0.060 kg is the mass

u = -32 m/s is the initial velocity

v = ? is the final velocity

By re-arranging the equation, we can find the final velocity of the ball after it hits the ground:

And the positive sign tells us that the direction of the velocity is upward (opposite to the initial velocity, which was downward)

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

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