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viva [34]
3 years ago
5

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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In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
cluponka [151]

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

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8 0
3 years ago
When you answer someones question how do you make it so your answer is verified??
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6 0
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A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
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Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

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Given;

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constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

v = u + at

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

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A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

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The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

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where

\rho=1025 kg/m^3 is the blood density

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So the pressure drop is:

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8 0
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