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.
Since force is mass*acceleration,
F = 70kg * 9.8 m/s
Answer:
Power is defined as Work / Time
P1 = P2 = W1 / T1 = W2 / T2 W1 and W2 are not necessarily the same Power expended depends on both work input over a unit of time
Answer:
Monitoring water quality is an important part of helping us determine whether or not we are making progress in cleaning up our waterways. It reveals the health and composition of streams, rivers, and lakes at a snapshot in time, as well as over weeks, months, and years. Dirty water could be dangerous to us, and our environment!
Answer:
<h3>
80000 loop needed to produce given induced emf.</h3>
Explanation:
Given :
Magnetic field
Area of coil
Average induced emf
The magnetic flux Φ =
Where No. of loop
From the faraday's law of electromagnetic induction,
Induced emf =
Where minus sign represent lenz law.
Here given in question , so
Induced emf =
Hence, 80000 loop needed to produce given induced emf.