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2 years ago

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Explanation: One can pick up a nail using the curved part of the horseshoe magnet farthest from the poles. A horse shoe magnet h

as magnetic fields all around it. In the presence of magnetic force fields, any magnetic object will be attracted to it.

1 answer:

PIT_PIT [208]2 years ago

4 0

Answer:

is there anymore choices than A

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Which of these is not true about the proton?

bulgar [2K]

Answer:

Option C is the untrue statement.

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3 years ago

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A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of

disa [49]

The average acceleration is

$\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}$

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A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

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3 years ago

An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnet

Vilka [71]

Answer:

Magnetic force is equal to $1.37\times 10^{-11}N$

Explanation:

We have given electron is accelerated with a potential difference of 81700 volt.

Magnetic field B = 0.508 T

Angle between magnetic field and velocity $\Theta =90^{0}$

Mass of electron $m=9.11\times 10^{-31}kg$

Charge on electron $e=1.6\times 10^{-19}C$

By energy conservation.

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700$

$v=169.4\times 10^6m/sec$

Magnetic force on electron

$F=qvBsin\Theta$

$F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}$

$=1.37\times 10^{-11}N$

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3 years ago

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Prove f=ma <br>i will give

iragen [17]

Please find attached photograph for your answer. Hope it helps. Please do comment

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