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masya89 [10]
3 years ago
14

Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito

r if the area of each plate is 5.5 cm2 ? express your answer using two significant figures.
Physics
1 answer:
Genrish500 [490]3 years ago
7 0
The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer
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A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about
Misha Larkins [42]
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

             1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is   

             (product of the 2 individuals) / (sum of the individuals)  .

In this problem, we have a humongous one and a tiny one.
Let's call them  1000  and  1 .
Then the series combination is

           (1000 x 1) / (1000 + 1)

        =       (1000) / (1001)

        =         0.999 000 999 . . . 

which is smaller than the smaller individual.

It'll always be that way.   </span>
5 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
A.
Elanso [62]

Answer:

Well, their are two answers in their. It would be Ask their Parent for assistance in persuading and Ask for an opportunity to earn extra credit:)

Explanation:

3 0
3 years ago
Read 2 more answers
PLEASE HELP Due today!
BigorU [14]
So i believe is exercise:)
7 0
3 years ago
I need help ASAPP
Tresset [83]

Answer: For the first part the answer is Radio Waves and the second part’s answer is gamma ray.

6 0
3 years ago
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