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3 years ago

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Two long parallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction. Find the magnitude of

the force per unit length that each wire exerts on the other wire and indicate if the force is attractive or repulsive.

1 answer:

olya-2409 [2.1K]3 years ago

5 0

To solve this problem we will apply the concepts related to electromagnetic force. This force is defined as the product between the free space permeability constant, the product of the two currents found, at the rate of the distance that separates them. In this way mathematically it can be given as,

$F = \frac{\mu_0 I_1I_2 }{2\pi d}$

Here,

$\mu_0$ = Permeability of free space

I = Each current

d = Distance

Replacing with our values,

$F = \frac{(4\pi *10^{-7})(5)(8)}{2\pi (0.3)}$

$F = 2.7*10^{-5} N$

Therefore the force is $2.7*10^{-5}N$

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Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides

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Answer:

B.the same as Lily's.

Explanation:

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3 years ago

The host says during the video that her interests during middle school were?

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Explanation:

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3 years ago

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.923 g, q = 4.52 µC is locat

Anestetic [448]

Answer:

$Q = -1.43\times 10^[-5} coulomb$

Explanation:

Given data:

particle mass = 0.923 g

particle charge is 4.52 micro C

speed of particle 45.7 m/s

In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve

$-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}$

solving for Q WE GET

$Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}$

$Q = -mv^2\times r \frac{4\pi \epsilon}{q}$

$Q = - \frac{0.923\times 10^{-3} \times 45.7^2\times (22.6\times 10^{-2})} {4.52\times 10^{-6} \times 9\times 10^9}$

where $\frac{1}{4\pi \epsilon} = 9\times 10^9$

$Q = -1.43\times 10^[-5} coulomb$

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4 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

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3 years ago

The engineering quadrangle is approximately 235 meters above sea level. If the temperature is 15 degrees Celsius, is that above,

RUDIKE [14]

Answer:

The 15 ⁰C measured at this altitude is above the standard temperature for the altitude.

Explanation:

The standard temperature at sea level is 15 degrees Celsius. It decreases about 2 degrees C (or 3.5 degrees F) per 1,000 feet of altitude above sea level.

235 meters is equal to 771 feet.

Using the formula below, we can estimate temperature loss due to this change in altitude, that is 771 feet above sea level.

temperature loss = (3.5 x Change in altitude)/1000ft

temperature loss = (3.5 x 771ft)/1000ft = 2.7⁰F, (32 -2.7 = 29.3 ⁰F)

this is equivalent to 1.5⁰C temperature loss.

Thus, the standard temperature of the engineering quadrangle at 235 meters above sea level is 13.5 ⁰C.

Therefore, the 15 ⁰C measured at this altitude is above the standard temperature for the altitude.

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4 years ago