Answer:
xaubUajnaai ajn AJ au aun a
Explanation:
ahayba uabah an aj
B is the correct answer hope that helped
The magnitude of the magnetic field inside the solenoid is 
.
The given parameters;
- <em>length of the solenoid, L = 91 cm = 0.91 m</em>
- <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
- <em>number of turns of the solenoid, N = 1300 </em>
- <em>current in the solenoid, I = 3.6 A</em>
The magnitude of the magnetic field inside the solenoid is calculated as;
where;
is the permeability of frees space = 4π x 10⁻⁷ T.m/A
Thus, the magnitude of the magnetic field inside the solenoid is .
Learn more here:brainly.com/question/17137684
We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
