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4vir4ik [10]
2 years ago
12

What potential difference vwc between the wire and the cylinder produces an electric field of 2.00Ã104 volts per meter at a dist

ance of 1.20 centimeters from the axis of the wire? (assume that the wire and cylinder are both very long in comparison to their radii.)?

Physics
1 answer:
vovikov84 [41]2 years ago
5 0
<span>The potential difference Vwc between the wire and the cylinder produces an electric field of 2.00Ã104 volts per meter at a distance of 1.20 centimeters from the axis of the wire is 1160 V</span>

In case of a long wire, electric field E is inversely proportional to r

Therefore E= x/r or x = E*r

Now, 

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A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
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Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

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Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

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2 years ago
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
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Answer:

Cis, Trans.

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5 0
2 years ago
The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the
irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

P_{abs}=P_{gauge}+P_{atm} (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

P_{gauge}=\rho gh (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh

P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)

P_{atm}=95950Pa

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

S.G=\frac{\rho_{fluid}}{\rho_{water}}

\rho_{liquid}=S.G*\rho_{water}

\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}

so:

P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa

P_{abs}=137642.5 Pa

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