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emmainna [20.7K]

2 years ago

11

Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw

o 4.0-cm-diameter disks.

1 answer:

aalyn [17]2 years ago

4 0

Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons

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A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$

$N_2 = 1400$

$\Delta V_{1rms} = 62V$

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

$\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}$

So the voltage for transformer 2 would be given by,

$\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}$

PART B) To express the number value we proceed to replace with the previously given values, that is to say

$\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}$

$\Delta V_{2rms} = \frac{1400}{160} 62V$

$\Delta V_{2rms} = 1446.66V$

7 0

2 years ago

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea

crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = $\frac{(172\ \mu C)}{V}$

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = $\frac{(172+220)\ \mu C)}{V}$

or

C' = $\frac{(392)\ \mu C)}{V}$

now the dielectric constant (k) is given as:

$k=\frac{C'}{C}$

substituting the values, we get

$k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}$

or

k = 2.279

6 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

8 0

2 years ago

How many dots should be drawn in the lewis dot structure for an atom of sodium A)19 B)8 C)4 D)1

Stells [14]

Na is in the first column on the periodic table so therefore it would have 1 valence electron
D 1

6 0

2 years ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

2 years ago