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True [87]
2 years ago
15

The compass is a small bar magnet that’s free to spin. Based on what you observed, is the coil of wire with current passing thro

ugh it a magnet?
The compass is a small bar magnet that’s free to spin. Based on what you observed, is the coil of wire with current passing through it a magnet?
Physics
1 answer:
OverLord2011 [107]2 years ago
6 0

Answer:

yes, it is a magnet.

Explanation:

The electrons that are travelling through the wire go in 1 direction allowing it to create a magnetic field around its area.

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Need help with a Physics math problem.
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Neither of those questions makes sense, and I believe that you should not waste your time worrying about them.

#61 gives you a lot of information about a ball, and then asks a question about a glove.

#62 gives a mysterious equation, doesn't tell you what either of the variables represents, and then asks for a quantity without ever telling us how that quantity is related to the equation.

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A rocket started from the rest and goes straight up with an acceleration of 4.0 m/s^2 for the first 100 seconds. Then it coasts
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Explanation:

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An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the
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The upward force is 737 N
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3 years ago
A 50 kg box hangs from a rope. what is the tension in the rope if
MrRa [10]
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3 years ago
Through what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinho
Mila [183]

Answer:

0.15 mV

Explanation:

In order to exhibit wave nature, the de Broglie wavelength of the electron must be of the same size of the diameter of the pinhole, therefore:

\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m

The de Broglie wavelength of an electron is

\lambda = \frac{h}{mv}

where

h=6.63\cdot 10^{-34} Js is the Planck constant

m=9.11\cdot 10^{-31} kg is the mass of the electron

v is the electron's speed

Therefore, the electron's speed must be

v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(1.0\cdot 10^{-7})}=7278 m/s

When accelerated through a potential difference \Delta V, the kinetic energy gained by the electron is equal to the change in electric potential energy, therefore

e\Delta V = \frac{1}{2}mv^2

where

e=1.6\cdot 10^{-19} is the magnitude of the charge of the electron

So, we can find the potential difference needed:

\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV

5 0
3 years ago
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