A0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. what is the mangnitude of the change in momentum of the stone?

change in momentum of the stone is 1.635 kg.m/s

Explanation:

let m = 0.500kg ball and M  be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone

the  initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J

the kinetic energy of the ball after rebounding is 70/100(100) = 70 J

Kb = 1/2mv^2

 v =  \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s  

from the conservation of linear momentum, we know that:

mvi + MVi = mvf + MVf

MVf - MVi = mvi - mvf

MVf - MVi = (0.500)(20) - (0.500)(16.73)

                 = 1.635 kg.m/s  

therefore, the change is momentum of the stone is 1.635 kg.m/s

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: