Answer:
L = 0.109 H
Explanation:
Given that,
Number of loops in the solenoid, N = 1500
Radius of the wire, r = 4 cm = 0.04 m
Length of the rod, l = 13 cm = 0.13 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
So, the self inductance of the solenoid is 0.109 henries.
Answer:h=5.81 m
Given
Mass of block(m)=0.250 kg
Spring Constant
Initial elongation =0.080 m=8 cm
Thus Initial Potential Energy stored =Final Potential Energy stored in Block
edge
- 256 lbs
The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero
; Then:
internal axial load at point D = Δ P
= -(P₂ - P₁)
= - ( 888 lbs - 632 lbs)
= - 256 lbs
7291.2 I hope this is right