Aheating element is made by maintaining a potential difference of 65.0 v across the length of a nichrome wire that has a 3.11 × 10–6 m2 cross section. nichrome has a resistivity of 5.00 × 10–7 ω·m. (a) if the element dissipates 6580 w, what is its length? (b) if 106 v is used to obtain the same dissipation rate, what should the length be?

a)l=3.99 m

b)l=10.57 m

Explanation:

Given that

V= 65 V

A= 3.11 x 10⁻⁶ m²

ρ=5 x 10⁻⁷  Ω·m

a)

P= 6580 W

We know that

P = V²/R

By putting the values

P = V²/R

6580 = 65²/R

R= 0.64 Ω

We know that

R= ρl/A

l= R.A/ρ

l= (0.64 x 3.11 x 10⁻⁶)/(5 x 10⁻⁷)

l=3.99 m

b)

V= 106 V

P = V²/R

6580 = 106²/R

R=1.70 Ω

l= R.A/ρ

l= (1.7 x 3.11 x 10⁻⁶)/(5 x 10⁻⁷)

l=10.57 m

f=ma newton

f=100x2.5=250n

conservation of mass

mass before something happens = mass afterwards