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3 years ago

OMG HELPPP PLS

Physics

1 answer:

Andre45 [30]3 years ago

7 0

It’s bad because the solar system controls the technology and affects your brain in a bad way

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

jok3333 [9.3K]

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }\\$ .......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$

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4 years ago

In the past, it was common for waste rocks from mines to be dumped in piles, known as tailings piles. These piles were sometimes

Tresset [83]

It increased deposition downstream.

7 0

3 years ago

Read 2 more answers

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

8 0

3 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

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4 years ago

Which of the following does not describe a scientific theory?

frez [133]

Explains how and why nature does what it does. ... A scientific law is a statement based on repeated experimental observations that ...

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3 years ago