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 1

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.


Related Questions

Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?

Answers

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

       

Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus

Answers

Answer:

2.55m

Explanation:

Using 1/do+1/di= 1/f

di= (1/f-1/do)^-1

( 1/0.0500-1/0.0510)^-1

= 2.55m

Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.

Answers

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.

Required:
In what directions is it possible that the wave is traveling?

Answers

Answer:

The wave is traveling in the y axis direction

Explanation:

Because the wave will always travel in a direction 90° to the magnetic and electric components

A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf

Answers

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = [tex]\pi r^{2}[/tex]

==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

[tex]I[/tex] = E/R

where R is the resistor

[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = [tex]I[/tex]E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.

Answers

Explanation:

It is given that,

Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.

Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]

As the buses are moving in opposite direction, then the concept of relative velocity is used. So,

Distance, [tex]d=v\times t[/tex]

v is relative velocity, v = 108 + 36 = 144 km/h

So,

[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]

So, the distance between them is 48 km after 20 minutes.

An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron

Answers

Answer:

acting force is the answer

The direction of the magnetic force on the moving electron is upward.

The direction of the magnetic force on the electron can be determined by applying right hand rule.

This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.

Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.

Learn more here:https://brainly.com/question/14434299

Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT

Answers

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

That is how:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

Answer:

[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]

Explanation:

[tex]\mathrm{force=mass \times acceleration}[/tex]

The force is given 0 newtons.

[tex]\mathrm{force=0 \: N}[/tex]

Plug force as 0.

[tex]\mathrm{0=mass \times acceleration}[/tex]

Divide both sides by mass.

[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]

[tex]\mathrm{0 =acceleration}[/tex]

[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]

Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?

Answers

Answer:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?

Answers

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

Calculation of the equivalent resistance:

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

learn more about resistance here: https://brainly.com/question/15047345

Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth

Answers

Answer:

Yes. Something similar occurs here on Earth.

Explanation:

Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.

Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine

A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev

Answers

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

     

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]

•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

Answers

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.

Answers

Answer:

206.67N

Explanation:

The sum of force along both components x and y is expressed as;

[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]

The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

[tex]a_x = \frac{d^2 x }{dt^2}[/tex]

[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]

Similarly,

[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]

[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]

[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?

Answers

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)

Answers

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.

Answers

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

[tex]distance=v\,*\, t[/tex]

[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister​

Answers

Current through 20 ohm resistor is 0.6 A

A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

Answers

Answer:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct

Answers

Answer:

B) the particle's momentum.

Explanation:

We know that

The centripetal force  on the particle when its moving in the radius R and velocity V

[tex]F_c=\dfrac{m\times V^2}{R}[/tex]

The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q

[tex]F_m=q\times V\times B[/tex]

At the equilibrium condition

[tex]F_m=F_c[/tex]

[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]

[tex]R=\dfrac{m\times V}{q\times B}[/tex]

Momentum = m V

Therefore we can say that the radius of curvature is directly proportional to the particle momentum.

B) the particle's momentum.

Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g

Answers

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.

Answers

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.

Answers

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10

Answers

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

Answers

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

 

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

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]

If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use

Answers

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?

Answers

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

Answers

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

Which has more mass electron or ion?

Answers

an ion has more mass than an electron
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