The sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)

2 x 10^10 years.

Explanation:

Given that the sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)

Let us first calculate energy from hydrogen gas.

4(1.007825) + 2(0.00549) - 4.002603 = 0.029795 amu

Since 4.0313 amu H+ > 0.029795 amu =  ratio of 1 to 0.00739

dE = ((2*10^30 kg) x 0.8 x 0.25)(3.00x10^8 m/s)^2 = 3.6x10^46 kg x m^2/s^2  

3.6x10^46 kg x m^2/s^2 x 0.00739 = 2.6604 x 60^44 kg x m^2/s^2

4x10^26 J/s x s = 2.6604x10^44 kg x m^2/s^2  solve for s.

s = 6.651x10^17 seconds

6.651x10^17 seconds x 1 min/60 s x 1 hr/ 60 min x 1 day/ 24 hr x 1 yr / 365 days = 2.1x10^10 years

 Based on the proposed overall fusion equation, it will therefore take the sun shine 2 x 10^10 years before it exhausts its hydrogen fuel.

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