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alexdok [17]
2 years ago
13

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

ne of charge has charge per unit length λ2 = -2.48 μC/m and is parallel to the x-axis at y1 = 0.400 mWhat is the magnitude of the net electric field at point y2 = 0.200 m on the y-axis?
Physics
1 answer:
Kitty [74]2 years ago
8 0

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

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8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

E=\frac{hc}{\lambda}

Where h corresponds to Plank constant (6.626070x10^-34Js), c is the speed of light in the vacuum (299792458m/s) and \lambda is the wavelength of the photon(in this case 800nm).

E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J

Tranform the units

1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV

The band Gap is 4eV, divide the band gap between the energy of the photon:

\frac{4ev}{1.549802445eV}=2.508974118

Rounding to the next integrer: 3.

Three photons are the minimum to equal or exceed the band gap.

4 0
3 years ago
In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el
kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

7 0
2 years ago
A student drops an object out the window of the top floor of a high-rise dormitory.
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-- The acceleration due to gravity is 32.2 ft/sec² .  That  means that the
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-- If dropped from "rest" (zero initial speed), then after falling for 4 seconds,
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-- 128.8 ft/sec = <em>87.8 miles per hour</em>

Now we can switch over to the metric system, where the acceleration
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-- Distance = (1/2) x (acceleration) x (time)²

       D = (1/2) (9.8) x (4)² =<em>  78.4 meters</em>

-- At 32 floors per 100 meters,  78.4 meters = dropped from the <em>25th floor</em>.


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3 0
2 years ago
Which state of matter takes both the shape and volume of its container?
Tresset [83]

The <em>gaseous state</em> of matter does that.  A gas expands to take the shape and volume of whatever you put it into.

6 0
3 years ago
What experiment should I make using Gravitational Force? <br><br>PLEASE HELP ME :)
Bogdan [553]

You could try the "Spinning Bucket" or the "Center Of Gravity" experiment. There are plenty more that you could research! Hope this helped :)

8 0
3 years ago
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