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Nady [450]
3 years ago
13

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

distance from the line will the field strength to be 2000 N/C?
a. D/√2.
b. √2D.
c. D/2.
d. 2D.
e. D/4.
Physics
1 answer:
Alla [95]3 years ago
5 0

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

E=\dfrac{k}{d^2}

We can write it as follows :

\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2

Put all the values,

\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}

So, the correct option is (a).

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Natasha2012 [34]

Answer:

0.117 m

Explanation:

First of all, we can find the wavelength of the wave in the problem, by using the wave equation:

v=f\lambda

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Solving for \lambda,

\lambda=\frac{v}{f}=\frac{350}{500}=0.7 m

This means that the distance between two consecutive points of the wave having a difference of phase of

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Here we want to find the distance between two points that have a difference of phase of

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d'=d\frac{\phi'}{\phi}=(0.7)\frac{\pi/3}{2\pi}=\frac{1}{6}(0.7)=0.117 m

7 0
3 years ago
Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve
dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

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h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

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ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

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H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

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q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

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Q = (23.46 x 10⁵ J)/1116 s

<u>Q = 20.22 x 10³ W = 20.22 KW</u>

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A cheetah is walking at 1.0 m/s when it sees a zebra 25 m away. what acceleration would be required to reach 20.0 m/s in that di
inna [77]
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8 0
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galina1969 [7]

Answer:

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Explanation:

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