A0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modulus of this wire?
a) 6.43 × 109 n/m2
b) 1.41 × 1010 n/m2
c) 5.66 × 1010 n/m2
d) 2.78 × 109 n/m2
e) 3.54 × 109 n/m2

(B) 1.41 x 10^{10} N/m^{2}

Explanation:

diameter of wire (d) = 0.6 mm = 0.0006 m

radius of wire (r) = 0.0006/2 = 0.0003 m

tension (force F) = 20 N

percentage strain = 0.5 %

what is the Young's modulus?

Young's modulus =

where

stress = force/area

        stress =

        stress =

        stress =70735530.3 N/m^{2}

strain = percentage strain ÷ 100

        strain = 0.5 ÷ 100 = 0.005

therefore Young's modulus = = = 1.41 x 10^{10} N/m^{2}

Multiply the 3 dimensions, assuming it is a rectangular cuboid, to get the volume in cm³ divide mass by volume to get density in g/cm³

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.