In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
where t is the time,
is the maximum charge on the capacitor at voltage V, and
is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using
and
, the time constant of the circuit is:
2) To find the charge on the capacitor at time
, we must find before the maximum charge on the capacitor, which is
And then, the charge at time
is equal to
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
. We can see this also from the previous formule, by using
:
Answer:
1.843 x 10^-5 C
Explanation:
<u><em>Givens:
</em></u>
It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.
Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation
E = (Q/A)/∈o (1)
Where,
Q: total charge on the disk.
A: the area of the disk.
<u><em>Calculations: </em></u>
We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold
Q = EA∈o
= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)
= 1.843 x 10^-5 C
note:
calculations maybe wrong but method is correct
<u>ALL of the following work assumes NO AIR RESISTANCE:</u>
1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²
2). a parabola
3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.
4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity
5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion
6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion
7). 45 m/s
8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand
9). a). 4.49 m/s; b). 29.7 m/s
10). 7.24 meters
11). 700 meters
12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters
600Hz is the driving frequency needed to create a standing wave with five equal segments.
To find the answer, we have to know about the fundamental frequency.
<h3>How to find the driving frequency?</h3>
- The following expression can be used to relate the fundamental frequency to the driving frequency;
f(n) = n * f (1)
where, f(1) denotes the fundamental frequency and the driving frequency f(n).
- The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.
f(4) = 4× f (1)
480 = 4× f(1)
f(1) = 480/4 =120Hz.
So, 120Hz is the fundamental frequency.
- To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
- For, n = 5,
f(n) = n 120Hz,
f(5) = 5×120Hz=600Hz.
Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.
Learn more about the fundamental frequency here:
brainly.com/question/2288944
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