Answer:
A. P = 18.75 watts
B. P = 75 watts
Explanation:
V = 120 Volts
P = VI
I = P/V = 75/120 = 0.625 Amps
V = IR
R = V/I
R = 120/0.625 = 192 Ω
So the resistance of the bulb is 192 Ω and it does not change as it is given in the question.
A. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 60 V?
As P = VI and I = V/R
P = V*(V/R)
P = V²/R
P = (60)/192
P = 18.75 watts
As expected, it will dissipate less power (18.75 watts) than rated power due to not having rated voltage of 120 Volts.
I = V/R = 60/192 = 0.3125 Amps
or I = P/V = 18.75/60 = 0.3125 Amps
Since the resistance is being held constant, decreasing voltage will also decrease current as V = IR voltage is directly proportional to the current.
B. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 120 V?
P = V*(V/R)
P = V²/R
P = 120/192 = 75 watts
I = P/V = 75/120 = 0.625 Amps
As expected, it will dissipate rated power of 75 watts at rated voltage of 120 Volts.
The main equivalency we can find that exists between the solid volume and liquid volume is that they both contain a definite volume. In addition, they are collectively called condensed phases due to the particles having virtual contact with each other.
Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
=4.826g/cm3.
In kg/m3,density=mass/volume. converting mass in grams to kg, 1000g=1kg,111g=0.111kg.
cm3 to m3, 1cm3=10^-6m3, 23cm3=0.000023m3
density=0.111kg/0.000023m3 or 2.3*10^-5=4,826.1kg/m3.
the other is a long process.