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SCORPION-xisa [38]
3 years ago
9

Problem 23.5 The electric field strength 1.5 cm from an electric dipole, on the axis of the dipole, is 1.5×105 N/C.

Physics
1 answer:
Nataly [62]3 years ago
3 0

We have that the dipole moment in nCm is considering the <em>parameters </em>

electric field strength 1.5 cm and  the axis of the dipole, is 1.5×105 N/C. is

  • P=112.5nCm

From the question we are told

The electric field strength 1.5 cm from an electric dipole, on the axis of the dipole, is 1.5×105 N/C.

<h3>Electric field</h3>

Generally the equation for the Electric field  is mathematically given as

E=\frac{2P}{4\pi*r^3}\\\\Therefore\\\\P=\frac{ 1.5×10^5*1.5*10^{-3}}{2}\\\\

  • P=112.5nCm

For more information on electric field visit

brainly.com/question/16517842

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5 0
4 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
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