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3 years ago

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An electric field of 2.09 kV/m and a magnetic field of 0.358 T act on a moving electron to produce no net force. If the fields a

re perpendicular to each other, what is the electron's speed?

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

The velocity is $v = 5838 \ m/s$

Explanation:

From the question we are told that

The electric field is $E = 2.09 kV/m = 2.09 *10^{3} \ V/m$

The magnetic field is $B = 0.358 \ T$

Generally the force experienced by the electron due to the magnetic field is

$F_m = qvB$

Generally the force experienced by the electron due to the electric field is

$F_e = qE$

Since from the question the net force is zero then

$F_e = F_m$

=> $v = \frac{E}{B}$

Substituting values

$v = \frac{2.09*10^{3}}{0.358 }$

$v = 5838 \ m/s$

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What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s

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4 years ago

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

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