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An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answers

Answer 1

Answer:

Answer:

(a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

$F = ma$ .....(I)

Using formula of electric force

$F = qE$ .....(II)

from equation (I) and (II)

$-qE= ma$

$a = (-qE)/(m)$

(a). We need to calculate the speed of the electron

Using equation of motion

$v = u+at$

Put the value of a in the equation of motion

$v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)$

$v=36813.18\ m/s$

$v =3.68*10^(4)\ m/s$

(b). We need to calculate the distance traveled by the electron

Using formula of distance

$s = ut+(1)/(2)at^2$

Put the value in the equation

$s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2$

$s=0.0000453\ m$

$s=4.53*10^(-5)\ m$

Hence, (a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

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You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle

Answers

Answer:

29.5 m/s

Explanation:

Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

Given Information:

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate = 0.111 liters/sec

Required Information:

Velocity = v = ?

Answer:

Velocity = 4.21 m/s

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

What is the most important safety rule to remember during lab activities

Answers

To follow instructions

g A particular guitar string has a length of 60.0 cm, and a mass per unit length of 2.00 grams/meter. You hear a pure tone of 660 Hz when a particular standing wave, represented by the animation above, is excited on the string. Calculate the wavelength of this standing wave.

Answers

Answer:

wavelength of the standing wave will be equal to 30 cm

Explanation:

We have given length of the guitar string L = 60 cm

Mass per unit length $\mu =2gram/m$

Frequency is given f = 660 Hz

We have to find the wavelength of the standing wave

Length of the string will be 2 times of the wavelength of the wave

So $L=2\lambda$

$\lambda =(L)/(2)=(60)/(2)=30cm$

So wavelength of the standing wave will be equal to 30 cm

HELP ME PLS!!!!Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge

Answers

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

What are cations and anions?

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

To learn more about the Cations and Anions:

brainly.com/question/980691

#SPJ5

Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer:

-9Q

Explanation:

Electric field at origin is:

$E=(2keQ)/(a^2)$

Electric field due to first charge at origin would be:

$E_1=(keQ)/(a^2)$

Electric field due to second charge would be:

$E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)$

If the second charge is Q', then $E_2$ should be:

$E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)$

compare the above two values to find the possible values of Q':

$(|Q'|)/(9)=Q\n |Q'|=9Q$

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.