Question

A metal wire 4m long has radius 0.5 mm the wire is suspended from a rigid support and a load of 10kg wt is attached at the other end what is the extension produced I'd posaiom ratio for the metal is 0.26 what is the lateral compression produced in it (y metal =1.25 × 10 10 N/m²)​

Answer 1

Answer:

hey mate

Explanation:

here is your answer

Data

L = 4 m

r = 0.5 m = 0.5 × 10-³m

g = 9.8 m/s²

posaions ratio of metal is 0.26

Y = 12.5 × 10 10 pascles

F = 10kg

at = 10 × 9.8

N = 98N

Solution

1.young modulus y = F/A /∆L/l = FL/A ∆ L

where L is the lenght of the wire

∆L =

$\frac{fl}{ay}$

= FL/πr²y

the extension produced

∆L = 9.8 × 4/3.142 × (0.5 × 10 -3)² × 12.5 × 10 1p

= 3.992 × 10 -3

2.possion ratio

posaions ratio = ∆d/d/∆L/L

= L/∆L× ∆d/d

where d is the diameter of the wire

lateral compression I e the change in the diameter of the wire

∆d = posaions ratio d ∆L/L

d = 2r = 2 × 0.5 × 10 -4 m

= 1 × 10 -3 m

∆d = 0.26 ×1 × 10 -³ × 3.992 × 10 -3 /4

= 2.595 × 10 -7 m

$thanku$

Answer 2

Data

L=4 m

r = 0.5

g = 9.8 m/s sq

posaions ratio of metal is 0.26

y=12.5×1010 pascles

F= 10 kg

at = 10×9.8

N = 98 N

Solution

1 young modulus Y = F/A/L /1 = FL/AL

where L is the length of the wire = f1/ay

= FL/pai r sq y

the large intention produce

L=9.8 × 4 /3.142 × (0.5 ×10^-3)^2×12.5×10 I p

= 3.992×10^-3

2. possion ratio

posaions ratio = d/d/ L/L

where d is the diameter of the wire

lateral compression I the change in the diameter of the wire

d = posions ratio

d = 2×0.5×10^-4

=1×10^-3

= 2.595×10^-7 m