The amount of charge on each plate is equal to 0.9 Coulombs.
<u>Given the following data:</u>
To calculate the amount of charge on each plate:
<h3>The formula for work done.</h3>
Mathematically, work done in moving a charge is given by this formula:
<u>Note:</u> The work done by an external force equals the change in potential energy for the charges that were moved.
But,
Now, the work done becomes:
Making q the subject of formula, we have:
Charge, q = 0.9 Coulombs.
Read more on charge here: brainly.com/question/14327016
Answer:
Position of the object at t = 5 s is centimeter
Explanation:
The position of object as a function of time can be represented as
It is given that at t seconds the position of the object is centimeter.
We will substitute these values to determine the value of
Velocity at initial stage when t seconds is
Position after seconds
Answer:
Explanation:
Heat generated by the coil is expressed as;
Heat generated = Power ×Time
Heat generated = IVt
I is the current = 4A
V is the voltage= 5V
t is the time = 4hrs
Convert to secs
t = 4(3600) = 14400secs
Substitute into the formula
Heat generated = 4×5×14400
Heat generated = 20×14400
Heat generated = 288000Joules
Heat generated =288KJ
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.