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3 years ago

What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?

Physics

2 answers:

aleksley [76]3 years ago

7 0

Answer:

Electric potential = V = 33 KV

Explanation:

The breakdown is an electric stress phenomenon. The value of electric field at which breakdown of dry air occur is 3 × 10^6 V/m. So,

E = 3 × 10^6 V/m

d = 1.1 cm = 1.1 × 10^-2 m

We know that:

E = V/d

V = Ed

V = (3 × 10^6)( 1.1 × 10^-2)

V = 33000 V

V = 33 KV

romanna [79]3 years ago

6 0

The electric field is given by volts/distance: $E= \frac{V}{d}$ . The breakdown voltage of dry air is about 3x10^6V/m. So solving for V we get
$V=Ed$
or $V=(3e6V/m)(0.011m)=33,000V$

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How does the energy moves from the suns core to the photosphere?

Minchanka [31]

Answer: Energy from the core travels by radiation through the radiative

zone, then by convection through the convection zone.

Explanation:

4 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

Which describes sublimation?

balu736 [363]

A. Solid turns into a gas

5 0

2 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago

When two or more capacitors are connected in series across a potential difference:

34kurt

Answer:

A) and B) are correct.

Explanation:

Let's take a look at the attached picture. Now

The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE

The equivalent capacitance is going to be: $\frac{1}{C_{total}}=\frac{1}{C_1} +\frac{1}{C_2}$

And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE

And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE

7 0

3 years ago