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2 years ago

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A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

nd is also at 20.0°C. Let Ecu be the electric field in the copper wire and Eag be the electric field in the silver wire. Use the resistivities at room temperature for copper. PCu = 1.72 x 10^-8.m and for silver PAE= 1.47 x 10^-8.m. What is the ratio Ecu/Eng?

1 answer:

IgorC [24]2 years ago

7 0

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

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Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

$f = \dfrac{nv}{2L}$

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

The length of the longer pipe is L = 2.30 m

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2 years ago

A student measures the speed of a rolling ball three times. She adds the measurements and divides by 3.What quantity did the stu

Vinvika [58]

Hey there!

A ball rolls 3 times. If she adds the three measurements and divides it by 3, it means that this is the average speed. The average speed is found by adding all measurements and dividing by the amount of numbers.

I hope this helps!

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2 years ago

Read 2 more answers

A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

irina [24]

Answer:

The atomic weight in g/mole of the metal (molar mass) is 8.87.

Explanation:

To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The chemical formula of metal oxide tells you that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

In the previous equation you can see that you need 3 oxygen anions to react with two metal cations. Then:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

You have 52.92 g of metal in the sample, then the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>

The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

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2 years ago

A projectile is launched at an angle of 60° from the horizontal and at a velocity of

gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

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2 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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2 years ago