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Luden [163]
3 years ago
11

Cynthia forgot to put the fabric softener in the wash. As her socks tumbled in the dryer, they became charged. If a small piece

of lint with a charge of +1.25 E - 10 C is attracted to the socks by a force of 3.0 E - 6 N, what is the magnitude of the electric field at this location?
Physics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

E = 24000 N/C = 24 KN/C

Explanation:

The electric field experienced by a test charge is given by the following formula:

E = \frac{F}{q}\\\\

where,

E = Electric Field = ?

F = Force of attraction = 3 x 10⁻⁶ N

q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C

Therefore, using these values in the equation, we get:

E = \frac{3\ x\ 10^{-6}\ N}{1.25\ x\ ^{-10}\ C}\\\\

<u>E = 24000 N/C = 24 KN/C</u>

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The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
solniwko [45]

The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

Wavelength=4*Length of the ear canal

As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

wavelength = 4*2.9*10^{-2} =0.116

Then the frequency is determined as

f=c/λ=339/0.116=2922 Hz=2.92 kHz.

So, the frequency of the human ear canal is 2.92 kHz.

4 0
3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Zarrin [17]

Answer:

\rm 9.186\times 10^{-7}\ C.

Explanation:

<u>Given:</u>

  • Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
  • Distance of separation between the plates, d = 1.0 cm = 0.01 m.
  • Minimum value of electric field that produces spark, \rm E=3\times 10^6\ N/C.

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

\rm E=\dfrac{\sigma}{\epsilon_o}.

where,

  • \rm \sigma = surface charge density of the plate of the capacitor = \dfrac qA.
  • \rm q = magnitude of the charge on each of the plate.
  • \rm A = surface area of each of the plate =\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.
  • \epsilon_o = electrical permittivity of free space, having value = 8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.

For the minimum value of electric field that produces spark,

\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0
3 years ago
ANSWER ASAP
shtirl [24]
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A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is
SOVA2 [1]
I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>
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gizmo_the_mogwai [7]
4m. The big travels 2m twice
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2 years ago
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