How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800 A

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

The inductance of the solenoid is given by;

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

a) 3.7 m/s^2

b) 231.8 N

Explanation:

Let m1 be mass of the first object (m1 = 38.0 kg) and let m2 be the mass of the second object (m2 = 17.0 kg ). Let a be the acceleration of the two objects. Let F1 be the force of gravity exerted on m1 and F2 be the force of gravity exerted on m2. Let M = m1 +m2

a)

F1 = m1g and F2 = m2g

So Fnet = F1 + F2

Since the pulleys will move in different directions when accelerating...

Fnet = F1 - F2

M×a = m1g - mg2

M×a = g×(m1 -m2)

a = g×(m1 - m2)/M

a = 9.8×(38 - 17)/(38 + 17)

a = 3.7 m/s^2

b)

Looking at the part for m2

Fnet  = T - m2g

-m2×a = T - m2g

T = m2(g - a)

T = 231.8 N