Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore = 1.482 ) and the cladding (ncladding = 1.44).
Suppose you wanted the largest angle at which total internal reflection occurred to be θmax = 5 degrees. What index of refraction does the cladding need if the core is unchanged?

Answers

Answer 1

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764


Related Questions

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time

Answers

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

Induced EMF:

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current

Answers

Answer:

Using ohms law

The current is found from Ohm's Law.

I = V /R = E /R = Bxy /Rt.

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction.
What is the Poynting vector S⃗ (x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction?

Answers

Given that,

The electric field is given by,

[tex]\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}[/tex]

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of [tex](\vec{E}\times\vec{B})[/tex] vector is the direction of propagation of the wave .

Direction of magnetic field = [tex]\hat{j}[/tex]

[tex]B=B_{0}\sin(kx-\omega t)\hat{k}[/tex]

We need to calculate the poynting vector

Using formula of poynting

[tex]\vec{S}=\dfrac{E\times B}{\mu_{0}}[/tex]

Put the value into the formula

[tex]\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}[/tex]

[tex]\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]

Hence, The poynting vector is [tex]\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]

what is rotation? a. the orbit of a satelite around a central body b. the motion of two objects around each other c the spinning of an object on its axis d the motion of an object around a central body

Answers

Answer:

C. The spinning of an object on its axis.

Explanation:

The definition of rotation answers this question.

A good way to remember is that when you rotate, you turn, and when you revolve, you move around.

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to:__________.
a) 4 Q.
b) 2 Q.
c) Q.
d) Q/2.
e) Q/4.

Answers

Answer:

D. Q/2

Explanation:

See attached file

A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb

Answers

Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

What is work?

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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A 22.0 cm diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?

Answers

Answer:

The average induced emf in the loop is 0.3 V

Explanation:

Given:

d = 22 cm

Magnetic field  is  B = 1.5 T

Change in time  Δt = 0.20 sec.

Radius of loop  = r = d/2 = 11 x [tex]10^{-2}[/tex] m

According to the faraday's law,  Induced emf is given by   e = - (ΔФ / Δt)

where Ф = magnetic flux

Ф = BAcos0   (in this occasion, Ф=0)

where area A = pi * r²

note that we have to neglect negative sign due to its lenz law...so

       B * pi * r

e = ----------------

          Δt

       1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²

e =  ------------------------------------

                        0.20

e = 0.3 V

   

Therefore, the average induced emf in the loop is 0.3 V

A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid

Answers

Answer:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

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The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?

Answers

Answer:

Torque = 8.38Nm

Explanation:

Time= 8.00s

angular speed (w) =400 rpm

Moment of inertia (I)= 1.60kg.m2 about its rotation axis

We need to convert the angular speed from rpm to rad/ sec for consistency

2PI/60*n = 0.1047*409 = 41.8876 rad/sec

What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

Then we need to use the formula below for our torque calculation

from basic equation T = J*dω/dt ...we get

Where : t= time in seconds

W= angular velocity

T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

Therefore, constant torque that is required is 8.38 Nm

Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.

Given-

Inertia of the flywheel is 1.60 kg m squared.

Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,

[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]

[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]

[tex]\omega = 41.89[/tex]

Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.

When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,

[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]

[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]

[tex]\tau=8.38[/tex]

Hence, the required constant torque is 8.38 N-m.

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How do you stay hydrated during warm-up and scheduled activity?

Answers

Answer:

In order to stay hydrated during warm-up(s) drink 8oz of water 20-30 mintues before you start exercising or during your warm-up(s), make sure you drink 7 to 10 oz of water every 10 to 20 minutes during exercise, and drink 8oz of water no more than 30 minutes after you exercise.

In order to stay hydrated during scheduled activity(s) drink 17 to 20 oz of water 2 to 3 hours before you start to exercise, like said before drink 8 oz of water 20 to 30 minutes before you start exercising or during your warm-up(s), drink 7 to 10 oz of water every 10 to 20 minutes during exercise, also said before drink 8 ounces of water no more than 30 minutes after you exercise.

Answer: My scheduled activity was one hour of softball practice. I play catcher, so my thighs and knees take a lot of abuse from kneeling and standing. The lunges were excellent at preparing my thighs for softball. The high knees exercise and arm pumping didn’t feed into softball too well. I suppose that they might help me with base running.

Explanation: EDMENTUM

A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion

Answers

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

Two copper wires have the same volume, but wire 2 is 10% longer than wire 1. The ratio of the resistances of the two wires R2/R1 is Group of answer choices 1.2. 1.1. 0.82. 0.91. 1.0.

Answers

Answer:

Explanation:

volume is same

π r₁² L₁  =π r₂²L₂

L₁ / L₂ = r₂² / r₁²

For resistance the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area

R₁ = ρ L₁ / S₁

R₂ = ρ L₂ / S₂

Dividing

R₁ / R₂ = L₁ / L₂ x S₂ / S₁

=  L₁ / L₂ x r₂² / r²₁

= L₁ / L₂ x L₁ / L₂

= L₁²/ L₂²

L₂ = 1.1 L₁ ( GIVEN )

= L₁²/ (1.1L₁)²

1 / 1.21

R₂ / R₁ = 1.21 .

= 1.2

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutual force between them

Answers

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]

Therefore, the mutual force between the two point charges is 319.64 N

A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:

Answers

Answer:

The expected year is 2017.

Explanation:

Total years that the millionaire to live = 15 years

Travel away from the earth at  = 0.8 c

This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:

[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]

Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017

At what temperature will water begin to boil and turn to steam?
212 degrees Celsius
100 degrees Fahrenheit
212 kelvins
100 degrees Celsius

Answers

Answer:

100 degrees Celsius

Explanation:

Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.

At 100 degrees Celsius water begin to boil and turns to steam.

What are the boiling point and melting point of water?

The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).

Is boiling water always 212?

If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.

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A cylindrical rod 120 mm long and having a diameter of 15 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which are possible candidates? Justify your choice(s).
Material Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson's ratio
Titanium alloy 70 250 0.33
Steel alloy 105 850 0.36
Magnesium alloy 205 550 0.27
Aluminium alloy 45 170 0.20

Answers

Answer:

The only suitable candidate is steel.

Explanation:

We are given;

Initial length;L_o = 120 mm = 0.12 m

Diameter;d_o = 15 mm = 0.015 m

Radius;r = 0.015/2 m = 0.0075 m

Applied force;F = 35000 N

Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm

Formula for Poisson’s ratio is;

v = ε_x/ε_z

Where;

ε_x is lateral strain = = Δd/d_o

ε_z is axial strain = σ/E

Thus;

v = (Δd/d_o)/(σ/E)

Making Δd the subject, we have;

Δd = (vσd_o)/E

Where σ is force per unit area

σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa

Now, let's find Δd for each of the metals given;

For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33

Thus;

Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm

Similarly, for steel;

Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm

For magnesium;

Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm

For aluminum;

Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm

Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.

Thus,the only possible candidate is stee.

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

Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec

Answers

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Velocity

Given :

h=1/2t²+1/2t

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

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What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same

Answers

Answer:

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N).
A) What is the magnitude of the component of Tom's weight parallel to the ladder?
B) What is the magnitude of the component of Tom's weight perpendicular to the ladder?

Answers

Answer: A) [tex]P_{x}[/tex] = 564.4 N

              B) [tex]P_{y}[/tex] = 374 N

Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:

[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]

[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]

side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]

side = 1.65

Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.

The image in the attachment illustrates the described above.

A) [tex]P_{x}[/tex] = P sen θ

[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]

[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]

[tex]P_{x}[/tex] = 564.4 N

B) [tex]P_{y}[/tex] = P cos θ

[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]

[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]

[tex]P_{y}[/tex] = 374 N

Two 2.0 g plastic buttons each with + 65 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to +250 nC. All three are released simultaneously.
a. How many interactions are there that have a potential energy?
b. What is the final speed of the left 2.0 g plastic button?
c. What is the final speed of the right 2.0 g plastic button?
d. What is the final speed of the 5.0 g plastic button?

Answers

Answer:

A) 3 interactions

B)  2.78 m/s

C)  2.78 m/s

D) 0 m/s

Explanation:

Given Data :

plastic buttons ( 2 ) = 2.0 g each

charge on each plastic buttons = +65 nC

A) The interactions that have potential energy can be expressed as

V net = v1 + v2 + v3

V net = 0 + [tex]\frac{kq1q2}{r12} + \frac{kq1q3}{r13} + \frac{kq2q3}{r23}[/tex]

hence the total interactions is (3)

B) The final speed of the left 2.0 g plastic button

U = [tex]\frac{9*10^9* 10^-18}{0.02} [ 65*250 + 65*250 + (65*65)/2 ][/tex]

U = 0.0155 J  

also U = [tex]2( 1/2 mv^2 )[/tex]

      U = mv^2 = 0.0155 J

therefore ;  v = [tex]\sqrt{0.0155/(2*10^-3)}[/tex] = 2.78 m/s

C) final speed of the right 2.0 g plastic button = Final speed of the right 2.0 g plastic button

v = 2.78 m/s

D) The final speed of the 5.0 g plastic button

= 0 m/s

An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=

Answers

Answer

0.2067m or 0.2067m

Explanation;

Let lenght of spring= Lo= 11cm=0.110m

It is hang from a mass of

3.05-kg having a length of L1= 12.40cm= 0.124m

Force required to stretch the spring= Fkx

But weight of mass mg= kx then K= Mg/x

K= 3.05-kg× 9.8)/(0.124m-.110m)

K=2135N

But potential Energy U= 0.5Kx

X=√ 2U/k

√(2*10)/2135

X=0.0967m

The required new length= L2= L0 ±x

=

.110m ± 0.0967m

X= 0.2067m or 0.2067m hence the total lenghth

What magnification in multiples is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed?

Answers

Answer:

51.6

Explanation:

A microscope is made up of a convex lens and the nature of the image formed by the object viewed from it is a virtual image. Since the image is virtual, the image distance will be negative and the focal length will be positive (for convex lenses).

Using the lens formula to first calculate the image distance from the lens;

1/f = 1/u + 1/v

f is the focal length = 0.150 cm,

u is the object distance = 0.155 cm

v is the image distance.

Since the image distance is negative,

1/f = 1/u - 1/v

1/0.150 = 1/0.155 - 1/v

1/v = 1/0.155 - 1/0.15

1/v = 6.542 - 6.667

1/v = -0.125

v = 1/-0.125

v = -8 cm

Magnification = image distance/object distance

Mag = 8/0.155

Mag = 51.6

Magnification produces is 51.6

The medical profession divides the ultraviolet region of the electromagnetic spectrum into three bands: UVA (320-420 nm), UVB (290-320 nm), and UVC (100-290 nm). UVA and UVB promote skin cancer and premature skin aging; UVB also causes sunburn, but helpfully fosters production of vitamin D. Ozone in Earth's atmosphere blocks most of the more dangerous UVC. Find the frequency range associated with UVB radiation.

Answers

Answer:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Explanation:

The frequency of an electromagnetic radiation can be given by the following formula:

υ = c/λ

where,

υ = frequency of electromagnetic wave = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = 290 nm to 320 nm

FOR LOWER LIMIT OF FREQUENCY:

λ = 320 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)

υ = 9.375 x 10¹⁴ Hz

FOR UPPER LIMIT OF FREQUENCY:

λ = 290 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)

υ = 10.34 x 10¹⁴ Hz

Therefore, the frequency range for UVB radiations is:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as originating from a point on the near side of the mirror.

Answers

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved  mirror than the centre of curvature.

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.

Answers

Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Wiley Coyote has missed the elusive road runner once again. This time, he leaves the edge of the cliff at 52.4 m/s with a purely horizontal initial trajectory. If the canyon is 141 m deep, how far from the base of the cliff does the coyote land

Answers

Answer:

280.86 m

Explanation:

Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;

x is the range

v is the velocity of the runner

t is the time taken

Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²

if u = 0

S = 1/2gt²

2S = gt²

t² = 2S/g

t = √2S/g

t = √2(141)/9.8

t = √282/9.8

t = 5.36secs

The range x = 52.4*5.36

x =280.86 m

Hence, the coyort lies approximately 280.86 m from the base of the cliff

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will have the electric field zero?​

Answers

Answer:

On that line segment between the two charges, at approximately [tex]0.7\; \rm m[/tex] away from the smaller charge (the one with a magnitude of [tex]5 \times 10^{-19}\; \rm C[/tex],) and approximately [tex]1.3\; \rm m[/tex] from the larger charge (the one with a magnitude of [tex]20 \times 10^{-19}\; \rm C[/tex].)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let [tex]k[/tex] denote the Coulomb constant, and let [tex]q[/tex] denote the size of a point charge. At a distance of [tex]r[/tex] away from the charge, the electric field due to this point charge will be:

[tex]\displaystyle E = \frac{k\, q}{r^2}[/tex].

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] and [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex]. Assume that the electric field is zero at [tex]r[/tex] meters to the right of the [tex]5\times 10^{-19}\; \rm C[/tex] point charge. That would be [tex](2 - r)[/tex] meters to the left of the [tex]20 \times 10^{-19}\; \rm C[/tex] point charge. (Since this point should be between the two point charges, [tex]0 < r < 2[/tex].)

The electric field due to [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}[/tex].

The electric field due to [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Note that at all point in this section, the two electric fields [tex]E_1[/tex] and [tex]E_2[/tex] will be acting in opposite directions. At the point where the two electric fields balance each other precisely, [tex]| E_1 | = | E_2 |[/tex]. That's where the actual electric field is zero.

[tex]| E_1 | = | E_2 |[/tex] means that [tex]\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Simplify this expression and solve for [tex]r[/tex]:

[tex]\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0[/tex].

[tex]\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0[/tex].

Either [tex]r = -2[/tex] or [tex]\displaystyle r = \frac{2}{3}\approx 0.67[/tex] will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, [tex]0 < r < 2[/tex]. Therefore, [tex](-2)[/tex] isn't a valid value for [tex]r[/tex] in this context.

As a result, the electric field is zero at the point approximately [tex]0.67\; \rm m[/tex] away the [tex]5\times 10^{-19}\; \rm C[/tex] charge, and approximately [tex]2 - 0.67 \approx 1.3\; \rm m[/tex] away from the [tex]20 \times 10^{-19}\; \rm C[/tex] charge.

In the figure, suppose the length L of the uniform bar is 3.2 m and its weight is 220 N. Also, let the block's weight W = 270 N and the angle θ = 45˚. The wire can withstand a maximum tension of 450 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?

Answers

Answer:

a)   x = 2.46 m

b)   318.2 N

c)    177.8 N

Explanation:

Need to resolve the tension of the string at end say B.

The vertical upward force at B due to tension is 450 sin 45°.  

Using Principle of Moments, with the pivot at A,

Anti clockwise moments = Clockwise moments

450 sin 45° X 3.2   = 220 X (3.2/2) + (270 X x)  

x = 2.46 m

(b) The horizontal force is only due to the wire's tension, so it is  

450 cos 45° = 318.2 N

(c) total downward forces = 270 + 220 = 496 N  

Total upward forces = 450 sin 45° (at B) + upForce (at A)

Equating, upForce = 496 - 318.2  

= 177.8 N

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?

Answers

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

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