Lets us consider an example:
Suppose a 10 ohm bulb is connected across the terminals of a 10 V
battery having 2ohm internal resistance.
Then total reistance in series we know, R1 + R2
Thus, R net = 10+ 2 = 12 ohm
The, current across circuit = 10/12= 0.833 A
Now, Power is given by
Thus, power dissipated across internal resistance, P =
And, total power dissipated =
Thsu, percentage of pwer not avaible = 1.37/8.2 = 16.70%
Answer:
The centripetal force on body 2 is 8 times of the centripetal force in body 1.
Explanation:
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :
Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :
So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.
Explanation:
1) N₂ + O₂ → 2 NO
Kc = [NO]² / ([N₂] [O₂])
Set up an ICE table:
Plug into the equilibrium equation and solve for x.
1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))
1.00×10⁻⁵ = (2x)² / (0.114 − x)²
√(1.00×10⁻⁵) = 2x / (0.114 − x)
0.00316 = 2x / (0.114 − x)
0.00361 − 0.00316x = 2x
0.00361 = 2.00316x
x = 0.00018
The volume is 1.00 L, so the concentrations at equilibrium are:
[N₂] = 0.114 − x = 0.11382
[O₂] = 0.114 − x = 0.11382
[NO] = 2x = 0.00036
2(a) Cl₂ → 2 Cl
Kc = [Cl]² / [Cl₂]
1.2×10⁻⁷ = (2x)² / (2 − x)
1.2×10⁻⁷ (2 − x) = 4x²
2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²
2.4×10⁻⁷ ≈ 4x²
x² ≈ 6×10⁻⁸
x ≈ 0.000245
2x ≈ 0.00049
2(b) F₂ → 2 F
Kc = [F]² / [F₂]
1.2×10⁻⁴ = (2x)² / (2 − x)
1.2×10⁻⁴ (2 − x) = 4x²
2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²
2.4×10⁻⁴ ≈ 4x²
x² ≈ 6×10⁻⁵
x ≈ 0.00775
2x ≈ 0.0155
F₂ dissociates more, so Cl₂ is more stable at 1000 K.