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iris [78.8K]
3 years ago
12

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

ciated with a uniform E field consist of a family of planes that are perpendicular to the E field.3-The electric potential difference between two locations can exist when there is only a source charge. Even when there is no TEST charge q, the potential difference can exist between two points in the space.
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

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Answer:

131.25

Explanation:

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What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H
Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 := [H^+]=0.01 mol/L

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Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

pH=-\log [H^+]

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2=-\log [H^+]\\

[H^+]=10^{-2}M= 0.01 M=0.01 mol/L

The hydrogen ion concentration at pH is equal to 6 = [H^+]

6=-\log [H^+]\\\\

[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L

Concentration of hydrogen ion at pH is equal to 2 =[H^+]=0.01 mol/L

Concentration of hydrogen ion at pH is equal to 6 = [H^+]'=0.000001 mol/L

The difference between hydrogen ion concentration at pH 2 and pH 6 :

= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L

Moles of hydrogen ion in 0.009999 mol/L solution :

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8 0
3 years ago
A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1
pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x   = D + (y/2)

1st speaker distance from x   = 1.80 + (0.513/2) = 2.0565 m   .....1

and

at 2nd speaker distance from x   = D - (y/2)

2nd speaker distance from x   = 1.80 - (0.513/2) = 1.5435 m     .........2

so destructive interference from 1 and 2  we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = (  0+ 0.5) wavelength

wavelength  = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd  min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = (  1 + 0.5) wavelength

wavelength  = 0.342 m

frequency 2 =  velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

7 0
3 years ago
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