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Gwar [14]
3 years ago
5

A pair of closely spaced parallel conducting plates, charged with equal and opposite electric charges, produces a uniform electr

ic field in the region between them. in designing a cutting‑edge device that will revolutionize the electronics industry, leticia sets up such a pair of plates separated by a distance 0.879 mm, then charges them so that the direction of the electric field in their interior region points from plate A to plate B. Her idea requires that electrons, when released from rest at one of the plates, reach the other plate at the speed of 1.97% of the speed of light. the speed of light is c = 3.00 x 10^8 m/s.
1. At which plate should the electrons be released?
O A
O B
O Either
O Cannot be determined
Physics
2 answers:
aleksandr82 [10.1K]3 years ago
6 0

Answer:

Plate B

Explanation:

Since the electric field is in the direction from plate A to plate B, this means plate A is the positive plate and plate B is the negative plate. The electron will flow in opposite direction to the direction of the electric field, meaning electrons will be released at B and accelerated toward plate A due to the attraction of the positive plate A on the negatively charged electron.

Sedbober [7]3 years ago
3 0

Answer:

Plate B.

Explanation:

If the direction of the electric field is from plate A to plate B, then this means that plate A is positively charged and plate B is negatively charged. If we are to move an electron between the plates, then we should place the electron on plate B, so the negatively charged electron can be attracted by the positive charges on plate A.

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KonstantinChe [14]

Answer:

Part A:

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=3.03*10^{-3} m^2/V

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

N-Au=\frac{Density*Avogadro Number}{atomic weight}

Density = 19.32g/cm^3

Avogadro Number=6.02*10^{23} atoms/mol

Atomic weight=196.97g/mol

So:

n=1.5*\frac{Density*Avogadro Number}{atomic weight}

n=1.5*\frac{19.32*6.02*10^{23}}{196.97}

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}

Electron Mobility=3.03*10^{-3} m^2/V

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4 years ago
In the chemical formula for an ionic compound, which item is written first?
Sav [38]

I’m not completely sure but if I had to guess I would say lithium oxide

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3 years ago
The Tesla Roadster electric car is rated at 16.8kW. How much energy does it use during a one hour drive? Answer in MJ
ASHA 777 [7]

(16.8 kW) means (16,800 joules/second)

Energy in 1 hour = (1 hour) x (16,800 J/s) x (3,600 s/hour)

Energy = (16,800 x 3,600) J

Energy =  60,480,000 Joules

Energy = 60.48 MJ

6 0
3 years ago
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You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using e
Mashutka [201]

You are working for a manufacturing company, which is mathematically given as

  • m=3\sqrt{2}
  • m=\frac{15\sqrt{5}}{16}
  • x=0.747a
  • m/n=\frac{(x^2+a^2)3/2}{x^3}

<h3>What is the value of m that will place the movable bead in equilibrium at x-a a ....?</h3>

a)

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3

m=3\sqrt{2}

b)

By force equilibrium

K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}

Therefore

n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\

m=\frac{15\sqrt{5}}{16}

c)

K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

d)

By force equilibrium

K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}

m/n=\frac{(x^2+a^2)3/2}{x^3}

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3 0
2 years ago
An object has a kinetic energy of 500 J and a mass of 30kg. how fast is the object moving?
yuradex [85]

Answer:velocity=5.8m/s

Explanation:

Mass(m)=30kg

Kinetic energy(ke)=500joule

velocity=√(2ke ➗ m)

Velocity=√(2(500) ➗ 30)

Velocity=√(1000 ➗ 30)

Velocity=√(100/3)

velocity=5.8m/s

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