7. If the forces shown in figure are in equilibrium find F1
and Fr
AF.
50N
30
37°
25N
53
25N
45°
F
40N
2012N
8. Find the tensions in the three strings. One string is horizontal, one
the other is inclined at 60° with the horizontal.
120
20 cm
Answers
Answer:
Answer:
Answer:
Answer: L=2m,
Answer: L=2m,d=3mm,A=
Answer: L=2m,d=3mm,A= 4
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .