*A 15000 kg railroad car travels on a horizontal track with a constant speed of 12 m/s. A 6000 kg load is dropped onto the car. What will be the car's speed?

8.57 m/s

Explanation:

From the question,

The total momentum of the car before the load was dropped = Total momentum of the call after the load was dropped.

Mu =  (M+m)V Equation 1

Where M = mass of the car, u = initial speed of the car, m = mass of the load, V = final speed of the load.

make V the aubject of the equation

V = mu/(M+m) Equation 2

Given: M = 15000 kg, m = 6000 kg, u = 12 m/s,

Substitute these values into equation 2

V = (15000×12)/(15000+6000)

V = 180000/21000

V = 8.57 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:

the answer would be   dignified - apex