The answer is D. Water breaking down into hydrogen and oxygen
HTH ^^
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.
Answer:
x = 16 [m]
Explanation:
This problem can be solved using the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration = 3 [m/s²]
t = time = 2 [s]
Now we can find the displacement using the following equation of kinematics.
Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,
P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi
Answer:
A.)
Explanation:
Because you didn't add anything or take anything away.