Abox slides from rest down a frictionless ramp inclined at 34.0° with respect to the horizontal and is stopped at the bottom of the ramp by a spring with a spring constant of k = 3.50 104 n/m. if the box has a mass of 12.0 kg and slides 3.00 m from the point of release to the point where it comes to rest against the spring, determine the compression of the spring when the box comes to rest.

x = 0.105 meters

Explanation:

It is given that,

A box slides from rest down a friction less ramp inclined at 34.0° with respect to the horizontal,

Spring constant of the spring,

Mass of the box, m = 12 kg

Th box slides 3 m from the point of release to the point where it comes to rest against the spring. We need to find the compression of the spring when the box comes to rest.

Let h is the height of the incline,

Using the conservation of energy in case of spring,

x = 0.105 meters

So, the compression of the spring is 0.105 meters. Hence, this is the required solution.

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:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j