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Bingel [31]
3 years ago
13

A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10

5 N ⋅ m 2 /C . What is the linear charge density (charge per unit length) on the rod?
Physics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

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What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?
vfiekz [6]

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

              L=10log_{10}\left ( \frac{I}{I_0}\right )

Here we need to find the intensity of sound (I).

               L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}

Substituting

          L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

          I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2

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8 0
3 years ago
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part t
UNO [17]

Answer:

\alpha=-3.01dB

Explanation:

From the question we are told that:

Sound level intensity

 \triangle I=40dB-80dB

Generally the equation for  intensity level  is mathematically given by

 \alpha=10log_{10}(I/I_x)dB

Where

 I= Intensity measured

 I_x=Threshold\ of\ audibility

 I_x= 10-12 W / m2

 \alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}

 \alpha= 10 log10 \frac{I_1}{I_2}

 \alpha=10 log10\frac{40}{80}

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Answer:

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T_{1} = Tension in string 1

T_{2} = Tension in string 2

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W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

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Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

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