Ask question

Ask question

All categories

strojnjashka [21]

3 years ago

7

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

article has a speed of 230 m/s and moves perpendicular to the magnetic field. Just as the particle enters the magnetic field, an electric field is turned on. What must be the magnitude of the electric field such that the net force on the particle is twice the magnetic force

1 answer:

miv72 [106K]3 years ago

3 0

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

<u>E = 420.9 N/C</u>

You might be interested in

What is the calculation for elastic potential?

grin007 [14]

Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.

3 0

3 years ago

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

3 years ago

What is the wavelength of a wave that has a speed of 160 m/s and a period of 3.7 ms?

Elza [17]

Answer:

160m/s

Explanation:

The speed of a wave is related to its frequency and wavelength, according to this equation:

v=f ×λ

3 0

2 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

<u>Where:</u>

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

$v$ = speed at which jumper jumps at all time

initial distance jumped is given as

$R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}$

final distance jumped is given as

$R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}$

Dividing final distance by initial distance

$\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}$

$\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}$

$R_{2} =7.38$

distance lost is given as

d = $R_{1} - R_{2}$

d = 7.4 - 7.38

d = 0.02 m

8 0

3 years ago