Give a rail gun shown below with b-3.0t and rail spacing of 25 cm. a discharging capacitor supplies a current of 100 ka. 5) a. what is the acceleration if the bar has a mass of 0.25 kg. [3 x 10s m/s? ] h. what is the bars speed at the end of a 2.75 m long track [1.2 km/s ]

Explanation:

It is given that,

Magnetic field, B = 3 T

Length of the spacing, L = 25 cm = 0.25 m

Current,

Mass of the bar, m = 0.25 kg

(a) Magnetic force is given by :

F = 75000 N

Using second law of motion,

F = m × a

(B) Let v is the speed of the bars at the end of long track. Initial speed of the bar is 0. Using third equation of motion as :

v = 1284.52 m/s

or

v = 1.2 km/s

So, the speed of the bar at the end is 1.2 km/s. Hence, this is the required solution.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.