what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates

Answers

Answer 1

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

             

the electric force is

             F_{e} = q E

   

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

         

               m = 2e E / g

     

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

 

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium


Related Questions

A positively charged particle has a velocity in the negative z direction at a certain point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
a. Bx is positive
b. Bz­ is positive
c. By is negative
d. By is positive
e. Bx is negative

Answers

Answer:

a. Bx is positive

Explanation:

See attached file

It took a student 30 minutes to drive from his home to campus on
Monday, and it took him 20 minutes on Tuesday driving the same
route. If on Monday he drove 36 mi/hr on average, what was his
average speed on Tuesday?
O 12 mi/hr
O 18 mi/hr
O 48 mi/hr
O 54 mi/hr
O 72 mi/hr​

Answers

Answer:

48 i believe

Explanation:

A flat slab of material (nm = 2.2) is d = 0.45 m thick. A beam of light in air (na = 1) is incident on the material with an angle θa = 46 degrees with respect to the surface's normal.
Numerically, what is the displacement, D, of the beam when it exits the slab?

Answers

Answer:

Explanation:

Formula of lateral displacement

[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]

t is thickness of slab , i  and r are angle of incidence and refraction respectively .

Given t = .45 m

sin i / sin r = 2.2

sin 46 / sin r = 2.2

sin r = .719 / 2.2 = .327

r = 19°

[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]

[tex]S_{lateral}=\frac{.45}{cos19} \times sin(46-19)[/tex]

= .45 x .454 / .9455

= .216 m

= 21.6 cm .

The displacement, D, of the beam when it exits the slab is; 21.65 cm.

We are given;

Refractive index of slab material; nm = 2.2

Thickness of slab; t = 0.45 m

Refractive index of air; na = 1

Angle of incidence; θa = 46°

From snell's law, we can calculate the angle of refraction from;

na × sin θa = nm × sin θm

Thus;

1 × sin 46 = 2.2 × sin θm

0.7193 = 2.2 × sin θm

sin θm = 0.7193/2.2

θm = sin^(-1) 0.32695

θm = 19.08°

Formula for the displacement of the beam is;

D = (t/cos θm) × sin (θa - θm)

Plugging in the relevant values gives;

D = (0.45/cos 19.08) × sin (46 - 19.08)

D = 0.4783 × 0.4527

D = 0.2165m = 21.65 cm

Read more at; https://brainly.com/question/24875145

A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).

Answers

Answer:

The value for  A  is A= 0.6

The angular acceleration is  [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]

Explanation:

From the question we are told that  

    The radius of the disk is  [tex]r = 25.0 \ cm = 0.25 \ m[/tex]

     The linear acceleration is  [tex]a(t) = At[/tex]

     At time   [tex]t = 3 \ s[/tex]

     [tex]a(3) = 1.80 \ m/s^2[/tex]

Generally angular acceleration is  mathematically represented as  

         [tex]\alpha(t) = \frac{a(t)}{r}[/tex]

Now  at t = 3 seconds  

         a(3) =  A *  3

=>      1.80 =  A  *  3  

=.>       A =  0.6

So  therefore

             a(t) =  0.6 t  

Now  substituting this into formula for angular acceleration

        [tex]\alpha (t) = \frac{0.6 t }{R}[/tex]

substituting for  r  

         [tex]\alpha (t) = \frac{0.6 t }{0.25}[/tex]

         [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]

 

     

A wire carries current in the plane of this screen toward the top of the screen. The wire experiences a magnetic force toward the right edge of the screen. Is the direction of the magnetic field causing this force

Answers

Answer:

The direction of the magnetic field causing this force is

In the plane of the screen and towards the bottom of the egde

Explanation:

This is by applying Fleming s right hand rule which explains that

When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction. The current in the wire can have two possible directions. Fleming's right-hand rule gives which direction the current flows.

The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other (at right angles), as shown in the diagram.[1]

The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.

The first finger is pointed in the direction of the magnetic field. (north to south)

Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)

Question 8 of 10
On which parts of the heating curve for water does adding thermal energy
mainly cause the particles to move faster?
200
150 -
B
To
100
Temperature ('C)
A
50
С
0
-50
10
40
50
60
70
Time (min)
O A. C and D
B. A and B
O O O O
O C. Band C
OD. B and D

Answers

Answer:

The correct answer is A    

Explanation:

In this exercise we are given a graph of temperature versus time.

In calorimeter processes there are two types

* one that when giving thermal energy to the system its temperature increases, this fundamentally due to the greater kinetic energy of the molecular ones, this process observes in the graphs as a straight line of constant slope

* A process donates all the thermal energy that is introduced is cracked in breaking the molecular bonds, taking matter from one thermodynamic state to another, for example: liquid to gas.

This process in curves as a horizontal line, that is, there is no temperature change,

When analyzing the graph shown, parts C and D are the one that show a change in temperature with thermal energy. The correct answer is A

Answer:

C and D

Explanation:

Just took the quiz

The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform E-field.

Answers

Answer:

The  magnitude of the  electric field intensity is  [tex]E = 7.89 *10^{6} \ V/m[/tex]

Explanation:

From the question we are told that

    The  voltage is  [tex]\epsilon = 72.7 \ mV = 72.7 *10^{-3} V[/tex]

    The  thickness of the membrane is  [tex]t = 9.22 \ nm = 9.22 *10^{-9} \ m[/tex]

     

Generally the electric field intensity is mathematically represented as

                [tex]E = \frac{\epsilon }{t}[/tex]

 substituting values

                [tex]E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}[/tex]

                [tex]E = 7.89 *10^{6} \ V/m[/tex]

If an object is placed at a distance of 10 cm in front of a concave mirror of focal length 4 cm, find the position and characteristics of the image formed. Also, find the magnification.

Answers

Answer:

Explanation:

Focal length f = - 4 cm

Object distance u = - 10 cm

v , image distance = ?

Mirror formula

[tex]\frac{1}{v} +\frac{1}{u} = \frac{1}{f}[/tex]

Putting the given values

[tex]\frac{1}{v} - \frac{1}{10} = - \frac{1}{4}[/tex]

[tex]\frac{1}{v}= - \frac{3}{20}[/tex]

v = - 6.67 cm .

magnification

m = v / u

= - 6.67 / - 10

= .667

so image will be smaller in size in comparison with size of object .

Characteristics will be that ,

1 ) it will be inverted and

2 ) it will be real image .

Search Results Web results A car of mass 650 kg is moving at a speed of 0.7

Answers

Answer:

W = 1413.75 J

Explanation:

It is given that,

Mass of car, m = 650 kg

Initial speed of the car, u = 0.7 m/s

Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s

Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]

So, the work done by the car is 1413.75 J.

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the electric field between the two plates?

Answers

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.

Answers

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you predict about the surface elevation for 50 km thick crust with an average density of 2.8 g/cm3

Answers

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw from the battery? Neglect any energy losses in getting energy from the battery to the motor.

Answers

Answer:

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

Explanation:

The formula of electric power is given as follows:

Power = (Voltage)(Current)

Current = Power/Voltage

In this question, we have:

Power = 45 KW = 45000 W

Voltage of Battery Pack = 340 V

Current needed to be drawn = ?

Therefore,

Current = 45000 W/340 V

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be

Answers

Answer:

The current will be 18 A

Explanation:

Given;

potential difference, V = 10 V

current between the resistor, I = 2 A

Apply ohm's law;

V = IR

R = V / I

R = 10 / 2

R = 5Ω

Resistance is given as;

[tex]R = \frac{\rho l}{A}[/tex]

where;

ρ is resistivity

l is length

A is area

[tex]R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}[/tex]

When the diameter of the resistor is tripled

d₂ = 3d₁

[tex]R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms[/tex]

The current is now calculated as;

Apply ohms law;

V = IR

I = V / R

I = 10 / 0.556

I = 17.99 A

I = 18 A

Therefore, the current will be 18 A

Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?

Answers

Answer:

The  wavelength is  [tex]\lambda = 1029 nm[/tex]

Explanation:

From the question we are told that

    The  wavelength of the left light is  [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]

      The  voltage across A  and  B is  [tex]V_{AB } = 1.2775 \ V[/tex]

Let the stopping potential  at A  be [tex]V_A[/tex] and the electric potential at B  be  [tex]V_B[/tex]

The voltage across A and B is mathematically represented as

      [tex]V_{AB} = V_A - V_B[/tex]

Now  According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

        [tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]

Where  W is the work function of the metal

             h is the Planck constant with values  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             c  is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

          [tex]eV_B = \frac{h * c}{\lambda } - W[/tex]

So  

      [tex]eV_{AB} = eV_A - eV_B[/tex]

=>    [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]

=>   [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]

=>   [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]

=>  [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]

Where e is the charge on an electron with the value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>   [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]      

=>  [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]  

=>   [tex]\lambda = 1.029 *10^{-6} \ m[/tex]

=>   [tex]\lambda = 1029 nm[/tex]

     

Two sound waves W1 and W2, of the same wavelength interfere destructively at point P. The waves originate from two in phase speakers. W1 travels 36m and W2 travels 24m before reaching point P. Which of the following values could be the wave length of the sound waves?
a. 24m
b. 12m
c. 6m
d. 4m

Answers

Answer:

a. 24 m

Explanation:

Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field

Answers

Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.8 AA , how many turns of wire would you need

Answers

Answer:

The number of turns of wire needed is 3536 turns.

Explanation:

Given;

length of the wire, L = 8 cm = 0.08 m

magnetic field on the wire, B = 0.1 T

current in the wire, I = 1.8 A

The magnetic field produced by a solenoid is calculated as;

B = μ₀ n I

where;

n is the number of turns per length = N / L

μ₀ is permeability of free space = 4π x 10⁻⁷ N/A²

[tex]B = \frac{\mu_o N I}{L} \\\\N = \frac{BL}{\mu_o I} \\\\N = \frac{0.1 *0.08}{4\pi*10^{-7} *1.8} \\\\N = 3536.32 \ turns[/tex]

Therefore, the number of turns of wire needed is 3536 turns.

Gravitational Force: Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one

Answers

Answer:

F' = F

Hence, the magnitude of the attractive force remains same.

Explanation:

The force of attraction between two bodies is given by Newton's Gravitational Law:

F = Gm₁m₂/r²   --------------- equation 1

where,

F = Force of attraction between balls

G = Universal Gravitational Constant

m₁ = mass of first ball

m₂ = mass of 2nd ball

r = distance between balls

Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:

F' = Gm₁'m₂'/r'²

here,

m₁' = 2 m₁

m₂' = 2 m₂

r' = 2 r

Therefore,

F' = G(2 m₁)(2 m₂)/(2 r)²

F' = Gm₁m₂/r²

using equation 1:

F' = F

Hence, the magnitude of the attractive force remains same.

What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  [tex]\lambda = 622 nm[/tex]

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  [tex]D = 5 \ m[/tex]

    The order of the fringe is m  =  6

     The distance between the slit is  [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]

    The fringe distance is  [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]

Generally the for a dark fringe the fringe distance is  mathematically represented as

        [tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]

=>     [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]

substituting values

=>      [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]

=>     [tex]\lambda = 6.22 *10^{-7} \ m[/tex]

       [tex]\lambda = 622 nm[/tex]

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial, the initial velocity is v0 and in the second it is 4v0. In the second trial, A : the amplitude is twice as great and the maximum acceleration is half as great. B : both the amplitude and the maximum acceleration are four times as great. C : the amplitude is half as great and the maximum acceleration is twice as great. D : both the amplitude and the maximum acceleration are twice as great. E : the amplitude is four times as great and the maximum acceleration is twice as great.

Answers

Explanation:

It is given that, in the first trial, the initial velocity is [tex]v_o[/tex] and in the second it is [tex]4v_o[/tex].

The total energy of the system remains constant. So,

[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=\text{constant}[/tex] ....(1)

x is amplitude

It means that the amplitude is directly proportional to velocity. If velcoity increases to four times, then the amplitude also becomes 4 times.

Differentiating equation (1) we get :

[tex]mv\dfrac{dv}{dt}+kx\dfrac{dx}{dt}=0[/tex]

Since,

[tex]\dfrac{dv}{dt}=a,\ \text{acceleration}[/tex] and [tex]\dfrac{dx}{dt}=v,\ \text{velocity}[/tex]

So,

[tex]mva+kxv=0[/tex]

It means that the acceleration is also proportional to the amplitude. So, acceleration also becomes 4 times.

Hence, the correct option is (B) "both the amplitude and the maximum acceleration are four times as great"

A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all

Answers

Answer:

The frequency will be reduced by a factor of √2/2

Explanation:

Pls see attached file

The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Let the initial mass of block be m.

And new mass is, 4m.

The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,

[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

Here,

k is the spring constant.

If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,

[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]

Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Learn more about the frequency of oscillation here:

https://brainly.com/question/14316711

The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air

Answers

Answer:

The critical angle is  [tex]i = 41.84 ^o[/tex]

Explanation:

From the  question we are told that

    The index of refraction of the sugar solution is  [tex]n_s = 1.5[/tex]

   The  index of refraction of air is  [tex]n_a = 1[/tex]

Generally from Snell's  law

      [tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]

Note that the angle of incidence in this case is equal to the critical angle

Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]

So  

      [tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]

      [tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]

      [tex]i = 41.84 ^o[/tex]

A spherical shell has inner radius 1.5 m, outer radius 2.5 m, and mass 850 kg, distributed uniformly throughout the shell. What is the magnitude of the gravitational force exerted on the shell by a point mass particle of mass 2.0 kg a distance 1.0 m from the center

Answers

Answer:

The magnitude of the gravitational force is 4.53 * 10 ^-7 N

Explanation:

Given that the magnitude of the gravitational force is F = GMm/r²

mass M = 850 kg

mass m = 2.0 kg

distance d = 1.0 m , r = 0.5 m

F = GMm/r²

Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.

F = (6.67 × 10^-11 * 850 * 2)/0.5²

F = 0.00000045356 N

F = 4.53 * 10 ^-7 N

An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 42 mm and 74 mm. The electric potential of the inner conductor with respect to the outer conductor is -308 V ( = 1/4πε0 = 8.99 × 10^9 N · m^2/C^2).

The maximum energy density of the capacitor is closest to:_______

Answers

Correct answer is 2.7 x 10^-3 J/m3

I hope that helps ! <33

The maximum energy density of the capacitor is closest to: 2.7 x 10^-3 J/m3.

What is meant by the energy density of a capacitor?

Energy density is defined as the total energy per unit volume of the capacitor. Since, Now, for a parallel plate capacitor, A × d = Volume of space between plates to which electric field E = V / d is confined. Therefore, Energy is stored per unit volume.

How do you calculate energy density?

All Answers (14) Energy density is equal to 1/2*C*V2/weight, where C is the capacitance you computed and V should be your nominal voltage (i.e 2.7 V). Power Density is V2/4/ESR/weight, where ESR is the equivalent series resistance.

Learn more about energy density at

https://brainly.com/question/13035557

#SPJ2

An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answers

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the total flux through the sphere are determined. Now the radius of the sphere is halved. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere

Answers

Answer:

The magnitude of flux remains the same, and the field increases.

Explanation:

This is because the number of field lines leaving the sphere remains constant and the electric field increases because the line density increases

If a 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft), how long could she power her 0.4 watt flashlight

Answers

Answer: 3217.79 hours.

Explanation:

Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).

Power = 0.4 watt

Mass of climber = 140 lb

= 140 x 0.4535 kg  [∵ 1 lb= 0.4535 kg]

⇒ Mass of climber (m) = 63.50 kg

Let [tex]h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ][/tex] and [tex]h_2= 4,600 ft = 1402.08\ m[/tex]

Now, Energy saved =[tex]mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J[/tex]

[tex]\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}[/tex]

Hence, she can power her 0.4 watt flashlight for 3217.79 hours.

ransverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What is the mass of the string

Answers

Answer:

0.055 kg

Explanation:

According to the given situation the solution of the mass of the string is shown below:-

Speed of the wave is

[tex]v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}[/tex]

[tex]30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}[/tex]

Mass of string is

[tex]= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}[/tex]

After solving the above equation we will get the result that is

= 0.055 kg

Therefore for calculating the mass of the string we simply applied the above formula.

The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass p=7.95 kg and radius p=0.89 m. The hanging masses are L=32.0 kg and R=17.8 kg. Calculate the magnitude of the masses' acceleration and the tension in the left and right ends of the rope, L and R , respectively.

Answers

Answer:

Acceleration(a) = 2.588 m/s²

TL = 230.784 N

TR = 220.5 N

Explanation:

Given:

M = 7.95 kg

mL = 32 kg

mR = 17.8 kg

g = 9.8 m/s²

Find:

Acceleration(a)

TL

TR

Computation:

Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]

Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]

Acceleration(a) = [139.16] / [53.775]

Acceleration(a) = 2.588 m/s²

TL = mL(g-a)

TL = 32(9.8-2.588)

TL = 230.784 N

TR = mR(g+a)

TR = 17.8(9.8+2.588)

TR = 220.5 N

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