Weak nuclear force is the most involved
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.
a). In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.
b). In series, the total current is V / (3R) .
In parallel, the total current is 3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.
c). In series, the power dissipated by the circuit is
(V) · V/3R = V² / 3R .
In parallel, the power dissipated by the circuit is
(V) · 3V/R = 3V² / R .
Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.
I think it’s “light energy to chemical energy” let me know if i’m wrong
Answer:
y = 6
Explanation:
Given parameters:
Given equation:
y = 2x + 10
And supposing; x = -2
Unknown:
y = ?
Solution:
We are going to solve this problem by substitution;
y = 2x + 10
since x = -2; input this into the equation to solve for y;
y = 2(-2) + 10 = -4 + 10 = 6
The value of y = 6
The boiling point of ethanol is at 78.37°C. So, the energy must include sensible heat to raise 19°C to the boiling point and latent heat to change liquid to gas. The equation would be
Energy = Sensible heat + Latent heat
Energy = mCpΔT + mΔH
For ethanol,
Cp = 46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ J/kmol·K
ΔH = 38,560 J/mol
Integrate the Cp expression to determine CpΔT:
CpΔT = ∫₂₉₂³⁵²(46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ )dT
The upper limit is (78.37+273) = 352 K, while the lower limit is (19 + 273) = 292.
CpΔT = 2384857192 J/kmol·K
2,000 J = m(2384857192 J/kmol)(1 kmol/1000 mol) + m(38,560 J/mol)
m = 8.253×10⁻⁴ moles of ethanol
Since the molar mass of ethanol is 46.07 g/mol,
Mass = (8.253×10⁻⁴ mol)(46.07 g/mol)
Mass = 0.038 g ethanol
