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Varvara68 [4.7K]
2 years ago
10

Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n

C charge.
The direction I am looking for is the angle that the resultant force makes with the positive x-axis. Do not include units for the angle and make sure you input answers for the angle in degrees.
Physics
1 answer:
bagirrra123 [75]2 years ago
6 0

Answer:

Bot Nm

Explanation:

jdjdmxjd mdjdcj jdsdj jedidj jddj

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Gamma ray technology can be used to do which of the following?
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The weight is not coming from the center of the mass because the force that act on it is not is equal is side.(2) section B donot have weight because the ruler bend down and section be raise up so no weight.
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3 years ago
Find the electron and hole mobilities, and the resistivity of intrinsic silicon at 300K. Is intrinsic silicon a semiconductor
tino4ka555 [31]

Answer:

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

Explanation:

Solution:

At 300K:

Let suppose mobility of electron in intrinsic semiconductor = M_{e}

Mobility of electron in intrinsic semiconductor is:

M_{e}  = 1300 cm^{2}/volt.sec

Let suppose mobility of hole in intrinsic semiconductor = M_{h}

M_{h} = 500 cm^{2}/volt.sec

We know that, intrinsic silicon semiconductor has equal number of holes and electrons. So,

At 300 K

Intrinsic Carrier Concentration = 1.5 x 10^{10}/cm^{3} = C

And,

Conductivity of intrinsic Silicon is:

σ = C x (M_{h} + M_{e}) e

e = 1.6 x 10^{-19} C

So, plugging in the values, we get:

σ = C x (M_{h} + M_{e}) e

σ = 1.5 x 10^{10} x (500 + 1300) x 1.6 x 10^{-19}

σ = 4.32 x 10^{-6}

So, now we can find the resistivity.

Resistivity = 1/σ

Resistivity = 1/ 4.32 x 10^{-6}

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

7 0
2 years ago
Assume that a lightning bolt can be modeled as a long, straight line of current. If 16 C of charge passes by a point in 1.50 x 1
Reptile [31]

Answer:

B = 7.9012*10^{-5}T

Explanation:

To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.

The current is defined by the load due to time, that is to say

I= \frac{q}{t}

Where,

q= Charge

t = time

So the current can be expressed as:

I = \frac{16}{1.5*10^{-3}}

I = 10666.67A

Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

B =\frac{\mu_0}{2\pi}\frac{I}{r}

Where,

\mu_0 =Permeability Constant

I= Current

r= radius

Replacing the values we have

B=\frac{4\pi*10^{-7}}{2\pi}(\frac{10666.67}{27})

B = 7.9012*10^{-5}T

7 0
3 years ago
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