Two identical 0.50-kg carts, each 0.10 m long, are at rest on a low-friction track and are connected by a spring that is initially at its relaxed length of 0.50 m and is of negligible inertia. You give the cart on the left a push to the right (that is, toward the other cart), exerting a constant 4.5-N force. You stop pushing at the instant when the cart has moved 0.55 m . At this instant, the relative velocity of the two carts is zero and the spring is compressed to a length of 0.30 m. A locking mechanism keeps the spring compressed, and the two carts continue moving to the right. By what amount do you change the system's kinetic energy?

The right answer will be "2.025 Joule".

Explanation:

According to the question,

The initial position of CM will be:

= 0.35 m

The final position of CM will be:

= 0.8 m

Now,

The CM moves to the right by:

=

=

The work done will be:

=

=

=

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: