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kipiarov [429]
3 years ago
5

The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are

Physics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

  • 0.829 = <em>x</em> / 10.8
  • <em>y </em>= 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

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Net charge produce due to presence of different chargers in the vicinity or in the considered region.

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Net electric charge is just a concept introduce to identify the result of the presence of number of charges which interact with each other. It can be found by treating electric chargers in algebraic form like numbers, where +charges get added with +charges

-charges get added with -charges

and when adding two charges with opposite signs they get subtracted and the symbol of the resultant charge comes from the charge with highest magnitude.

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The diagram shows an LED light bulb.
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How can you make the moon go around in a bigger circle
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You would have to give it more mechanical energy.

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Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l
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The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
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