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harkovskaia [24]
3 years ago
14

Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric fiel

d on the y axis at y
Physics
1 answer:
ad-work [718]3 years ago
5 0

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

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Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

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Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

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Hence, The value of the average convection coefficient is 20 W/Km².

7 0
3 years ago
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