Ask question

Ask question

All categories

3 years ago

12

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

traveling a distance d.
a. What is the magnitude of the electric field?
b. What is the direction of the electric field?

1. in the direction of the electron's motion
2. opposite to the direction of the electron's motion
3. perpendicular to the direction of the electron's motion

1 answer:

KIM [24]3 years ago

7 0

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = $\frac{1}{2}mv_f^2$

But the final velocity is 0 (at rest). Hence:

Final KE = $\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0$

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion

You might be interested in

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly

Luba_88 [7]

Answer:

It falls at the same speed in both cases.

Explanation:

If I were standing still the phone would be in free fall after slipping out of my hand.

I set a frame of reference with origin on the ground and the positive Y axis pointing up.

It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.

It would be subject to an gravitational acceleration of -32.2 ft/s^2.

Since acceleration is constant:

Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2

When it hits the floor at t1 it will be at Y(t1) = 0

0 = 5 + 0 * t1 - 16.1 * t1^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

If the elevator is standing still it would take 0.55 s to hit the ground.

Now, if the elevator is moving up at 10 ft/s.

The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t

Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.

And it will hit the floor of the elevator not at 0, but at

Ye = 10 * t2

So:

10 * t2 = 5 + 10 * t2 - 16.1 * t2^2

0 = 5 - 16.1 * t2^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

$t1 = \sqrt{0.31} = 0.55 s$

It falls at the same speed in both cases.

4 0

3 years ago

Las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si

Sveta_85 [38]

La altura es de 169.4 metros.

Dado que las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si el cable tensor más corto mide 10m, para determinar cuál es altura de un cable que se encuentra a 100m de distancia del centro se debe realizar los siguientes cálculos, aplicando la ecuación parabólica:

(240)² = 4P x (110-10)
57600 = 4P x 100
57600 = 400P
57600/400 = P
144 = P
200 x 200 = 4 x 144 x (Altura - 100)
40000 = 576Altura - 57600
40000 + 57600 / 576 = Altura
169.4 metros = Altura

Por lo tanto, la altura es de 169.4 metros.

Aprende más en brainly.com/question/20333463

8 0

3 years ago

A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?

seraphim [82]

Answer:

No.

Explanation:

$P_r$ = Power rating = 1 W

R = Resistance = $350\ \Omega$

V = Voltage = $24\ \text{V}$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}$

$1.65\ \text{W}>1\ \text{W}$

So

$P>P_r$

Hence, the resistor is not operating within its power rating.

8 0

3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 $\frac{m^{3} }{s}$ </h3>

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago