Answer:
D The temperature, pressure and volume will all increase
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
Answer:
relative density is the density divided by some reference density.
for liquids like mercury the reference density is most likely water, so
relative density = density mercury / density of water = 13600
Answer:
Yes it is possible to increase the power with out changing the amount of work.
Explanation:
The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.
Work = 500 [J]
Time = 5 [s]
Power will be:
![Power=\frac{500}{5} \\Power=100 []watt]\\](https://tex.z-dn.net/?f=Power%3D%5Cfrac%7B500%7D%7B5%7D%20%5C%5CPower%3D100%20%5B%5Dwatt%5D%5C%5C)
Now if we change the time to 2 seconds:
![Power = 500 [J]/2[s]\\Power = 250 [watt]\\](https://tex.z-dn.net/?f=Power%20%3D%20500%20%5BJ%5D%2F2%5Bs%5D%5C%5CPower%20%3D%20250%20%5Bwatt%5D%5C%5C)
As we can see, the power was increased without the need to change the work.
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