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3 years ago

10

If a certain quantity of gas has a volume of 30cm cube day a pressure of 110^5 pa what is its volume when the pressure is 210^

5

1 answer:

Softa [21]3 years ago

4 0

Answer:

200000

Explanation:

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The vanishing of an ionic solid (like table salt) would be an example of acting like a solvent

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What accurately describes what happens when water vapor condenses into dew in terms of energy

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The awnser is condensation

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What is the atomic mass of an element?

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3 years ago

Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

$\dfrac{F}{L}= \dfrac{KI_1I_2}{d}$

$I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}$

$I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}$

I₂ = 32.67 A

distance where the magnetic field is zero

$\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}$

$y_1 = 0.248\ m$

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0

3 years ago

Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al

Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire $(m) = 1.10g = 1.10*10^{-3} kg$

Density $(\rho)= 8.92*10^3kg/m^3$

Resistively of copper $(\gamma) = 1.7*10^{-8}\Omega \cdot m$

Resistance (R) = 0.390\Omega

Volume is defined as,

$V= lA \text{ and }\frac{m}{\rho}$

$lA= \frac{1.10*10^{-3}}{8.92*10^3}$

$lA = 1.233*10^{-7} m^3$ (1)

We know that,

$\frac{l}{A} = \frac{R}{\gamma}$

$\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}$

$\frac{l}{A} = 2.2941*10^7 m^{-1}$ (2)

Multiplying equation we have

$l^2 = (1.233*10^{-7})( 2.2941*10^7)$

$l^2 = 2.8286m^2$

$l =\sqrt{2.8286m^2}$

$l = 1.68m$

Therefore the length of the wire is 1.68m

6 0

3 years ago