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eimsori [14]
3 years ago
11

A long, straight metal rod has a radius of 5.70 cm and a charge per unit length of 28.2 nC/m. Find the electric field at the fol

lowing distances from the axis of the rod, where distances are measured perpendicular to the rod's axis. (a) 3.10 cm magnitude N/C direction (b) 18.0 cm magnitude N/C direction(c) 180 cm magnitude N/C direction
Physics
1 answer:
tekilochka [14]3 years ago
5 0

Answer:

0

2817.43213 N/C

281.74321 N/C

Explanation:

The metal rod is conducting electricity. This means that the charges are being carried on the surface. Inside the rod there is no electric field. So, the electric field intensity is zero at a distance 3.1 cm.

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

r = Distance

\lambda = Charge per unit length = 28.2 nC/m

Electric field is given by

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 0.18}\\\Rightarrow E=2817.43213\ N/C

The electric field intensity is 2817.43213 N/C

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 1.8}\\\Rightarrow E=281.74321\ N/C

The electric field intensity is 281.74321 N/C

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Answer:

The color you get for mixing red and yellow is orange.

Explanation:

Minecraft logic

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3 years ago
Fill in the term that best completes the statement.<br> Work transfers<br> between objects.
n200080 [17]

Answer:

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<em><u></u></em>

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7 0
4 years ago
Determine the rate at which the electric field changes between the round plates of a capacitor, 5.8 cm in diameter, if the plate
Natalka [10]

Answer:

The rate at which the electric field changes between the round plates of a capacitor is 125\times 10^{3}Vs^{-1}.

Explanation:

It is given in the problem that the round plates of a capacitor are spaced some distance apart and the voltage across them is changing.

The expression for the electric field in terms of voltage is as follows;

E=\frac{V}{d}

Here, E is the electric field, V is the voltage and d is the distance of separation.

Differentiate expression of the electric field with respect to time, t.

\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}

Convert the distance of separation from mm to m.

d= 1.2 mm

d=1.2\times 10^{-3}m

Calculate the rate at which the electric field changes.

\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}

Put \frac{dV}{dt}=150 Vs^{-1} and d=1.2\times 10^{-3}m

\frac{dE}{dt}=\frac{1}{1.2\times 10^{-3}}(150)

\frac{dE}{dt}=125\times 10^{3}Vs^{-1}

Therefore, the rate at which the electric field changes is 125\times 10^{3}Vs^{-1}.

4 0
3 years ago
Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?
Norma-Jean [14]

Answer:

Part a)

\lambda = 300 m

Part b)

E = 2.7 N/C

Part c)

I = 9.68 \times 10^{-3} W/m^2

P = 3.22 \times 10^{-11} N/m^2

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

\lambda = \frac{c}{f}

\lambda = \frac{3\times 10^8}{1\times 10^6}

\lambda = 300 m

Part b)

As we know the relation between electric field and magnetic field

E = Bc

E = (9 \times 10^{-9})(3\times 10^8)

E = 2.7 N/C

Part c)

Intensity of wave is given as

I = \frac{1}{2}\epsilon_0E^2c

I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)

I = 9.68 \times 10^{-3} W/m^2

Pressure is defined as ratio of intensity and speed

P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}

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6 0
3 years ago
A horizontal pipe carrying water tapers from 200mm diameter
Tomtit [17]

Answer:

Delta pressure = 45938[Pa]

Explanation:

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Now we can calculate the two velocites VA and VB

Using the bernoulli equation we have:

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Where ro = density of water = 1000[kg/m^3]

P_{B}-P_{A}= 1000*(7^2-1.75^2)\\P_{B}-P_{A}= 45937.5[Pa]

5 0
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