Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of 100 kg Young's modulus of wire is 12.5 x 1010 Nm-2.

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j