Explanation:
30 minutes is 1800 seconds.
Power = energy / time
P = 1100 J / 1800 s
P = 0.611 W
Converting the energy from J to kWh:
1100 J × (1 Ws / 1 J) × (1 kW / 1000 W) × (1 h / 3600 s) = 3.06×10⁻⁴ kWh
To find resistance, you need to be given either the voltage or the current.
Yeah!! It is better to decrease the speed of the vehicle so the ejection of gases would happen in slow manner.... then you can save your gasoline
Answer:
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,
- separation distance of capacitor 1,
- quantity of charge on capacitor 2,
- quantity of charge on capacitor 1,
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
From eq. (1)
For capacitor 2:
For capacitor 1:
![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
& for capacitor 1:
![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
