Change in speed = (acceleration) x (time)
4 minutes = 240 seconds
Change in speed = (40 m/s²) x (240 seconds)
Change in speed = <em>9,600 m/s</em>
What you're actually describing here is a car pulling 4 G's for 4 minutes, and ending up going 21,475 miles per hour.
The driver would definitely NOT get a speeding ticket, because nobody could catch him.
Also, his car would heat up and shoot flames from atmospheric friction.
(He could avoid this with some fancy steering, leave the atmosphere, and end up in low-Earth-orbit.)
Actually, I hope there's nobody in the car. His experience wouldn't be pretty.
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
Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
![V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V](https://tex.z-dn.net/?f=V%3Dk%5B%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%5D%5C%5C%5C%5CV%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0125m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.018m%7D%5D%3D1.6%2A10%5E6V)
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
![V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V](https://tex.z-dn.net/?f=V%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0025m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.033m%7D%5D%3D13.35%2A10%5E6V)
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
Answer:
Explanation:
A and B are touched .
charge on each of them after touching = (9q - q) / 2 = 4q
when C is touched with A
charge on A and C each after touching
= 4q + 0 / 2 = 2q
When C is touched with B
charge on each of them
(2q + 4q ) / 2 = 3 q
Final charge on C = 3q
