Physics, 15.04.2020 00:55 AdoNice

Now assume that the charge is emitted with velocity v0 in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x direction. What should the magnitude E of the electric field be if the charge is to hit the target on the screen?

Answers: 3

Show answers

Answers

Answer from: MarishaTucker

Complete Question:

An charge with mass mand charge qis emitted from the origin, (x,y)=(0,0) . A large, flat screen is located at x=L. There is a target on the screen at y positiony_h, where y_{\rm h} /> 0 . In this problem, you will examine two differentways that the charge might hit the target. Ignore gravity in thisproblem.

a.)

Assume that the charge is emitted withvelocity v_0 in the positive x direction. Between the originand the screen, the charge travels through a constant electricfield pointing in the positive y direction. What shouldthe magnitude Eof the electric field be if the charge is to hit the target on thescreen?

Express your answer in terms ofm,q,y_h, v_0, and L.

b.)

Now assume that the charge is emitted withvelocity v_0 in the positive y direction. Between the originand the screen, the charge travels through a constant electricfield pointing in the positive x direction. What shouldthe magnitude E of the electric field be if the charge is to hit the target on thescreen?

Express your answer in terms ofm,q,y_h, v_0, and L.

a) $E = \frac{mv_{0}^{2} y_{h} }{0.5qL^{2} }$

b) $E = \frac{mv_{0}^{2} L }{0.5qy_{h} ^{2} }$

Explanation:

The velocity of charge in the x direction = v₀

The time required by the charge to hit the screen at a distance L,

t = L/v₀

The force on the charge, F = qE

F = ma

Equating the two relations for force

ma = qE

acceleration of the charge, a = qE/m

The initial velocity of the charge along the y-direction is o, the vertical distance covered by the charge is:

y = 0.5at²

substituting the relations for a and t

y = 0.5* (qE/m)*(L/v₀)²

$y = \frac{0.5qL^{2} E}{mv_{0}^{2} }$

If the charge is to hit the target on the screen $y = y_{h}$

$y_{h} = \frac{0.5qL^{2} E}{mv_{0}^{2} }$

Making E the subject of the formula, the magnitude of the electric field is:

$E = \frac{mv_{0}^{2} y_{h} }{0.5qL^{2} }$

The velocity of charge in the y direction = v₀

The time required by the charge to hit the screen at a height $y_{h}$ ,

$t = \frac{y_{h} }{v_{0} }$

The force on the charge, F = qE

F = ma

Equating the two relations for force

ma = qE

acceleration of the charge, a = qE/m

The initial velocity of the charge along the x-direction is o, the horizontal distance covered by the charge is:

x(t) = 0.5at²

substituting the relations for a and t

$x(t) = 0.5(qE/m)(y_{h}/v_{0} ) ^{2}$

$x(t) = \frac{0.5qy_{h} ^{2} E}{mv_{0}^{2} }$

If the charge is to hit the target on the screen x(t) = L

$L = \frac{0.5qy_{h} ^{2} E}{mv_{0}^{2} }$

Making E the subject of the formula, the magnitude of the electric field is:

$E = \frac{mv_{0}^{2} L }{0.5qy_{h} ^{2} }$

Answer from: Quest

Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy

Answer from: Quest

answer: b.

explanation:

Another question on Physics

Physics, 21.06.2019 19:30

Energy is the ability to do work which measures what

Answers: 3

Answer

Physics, 22.06.2019 08:30

Hey student studies gravity using objects that have the same mass which two objects have the greatest gravitational force acting between them a. 100kg 1.0m 100kg b. 100kg 2.0m 100kg c. 100kg 2.0m 100kg big d. 100kg big 3.0m 100kg big

Answers: 1

Answer

Physics, 22.06.2019 16:30

The total amount of energy contained in an object

Answers: 3

Answer

Physics, 22.06.2019 18:50

8.29 two streams containing pyridine and acetic acid at 25°c are mixed and fed into a heat exchanger. due to the heat-of-mixing effect, it is desired to reduce the temperature after mixing to 25°c using a stream of chilled ethylene glycol as indicated in the diagram. calculate the mass flow rate of ethylene glycol needed. the heat capacity of ethylene glycol at these conditions is approximately 2.8 kj/(kg k), and the enthalpy change of mixing (δmixh) is given below.

Answers: 3

Answer

You know the right answer?

Now assume that the charge is emitted with velocity v0 in the positive y direction. Between the orig...

Questions