Answer:
The percentage is [tex]k = 12.5 \%[/tex]
Explanation:
From the question we are told that
The axis is is at [tex]\theta = 45 ^o[/tex]
Generally the of intensity light emerging from the first polarizer is mathematically represented as
[tex]I_{1} = \frac{I_o}{ 2}[/tex]
Where [tex]I_o[/tex] is the intensity of unpolarized light
Now the light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]
[tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]
[tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]
Now the light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]
[tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Now the percentage of the intensity of light that emerged with respect to the intensity of the unpolarized light is
[tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]
[tex]k = 12.5 \%[/tex]
The percentage of light that gets through the three successive Polaroid filters is; 12.5%
We are given;
Angle of transmission axis; θ = 45°
Formula for intensity of light from first polarizer is;
I₁ = ¹/₂I₀
Formula for intensity of light from second polarizer is;
I₂ = I₁cos²θ
Formula for intensity of light from third polarizer is;
I₃ = I₂cos²(90 - θ)
Combining the 3 equations;
Put ¹/₂I₀ for I₁ in second formula to get;
I₂ = ¹/₂I₀cos²θ
Put ¹/₂I₀cos²θ for I₂ in third formula to get;
I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)
Plugging in 45° for θ gives;
I₃ = ¹/₂I₀cos²45*cos²(90 - 45)
⇒ I₃ = ¹/₂I₀cos²45*cos²45
⇒ I₃ = ¹/₂I₀cos⁴45
Now, cos 45 in surd form is 1/√2. Thus;
I₃ = ¹/₂I₀(1/√2)⁴
I₃ = ¹/₂I₀(¹/₄)
I₃ = ¹/₈I₀
I₃/I₀ = ¹/₈
I₃/I₀ = 0.125
In percentage form, we have;
I₃/I₀ = 12.5%
Read more about unpolarized light at; https://brainly.com/question/1444040
A lamp has the shape of a parabola when viewed from the side. The lamp is centimeters wide and centimeters deep. How far is the light source from the bottom of the lamp if the light source is placed at the focus
The question is not complete so i have attached it.
Answer:
The light source is 2 cm from the bottom of the lamp
Explanation:
From the attached image, we can see that the parabola opens up with its vertex at the origin.
Now, the standard form of equation for a parabola is:
x² = 4ay
Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)
Thus, we have;
12² = 4a(18)
144 = 72a
a = 144/72
a = 2
Looking at the parabola again, the line of symmetry is at x = 0
Thus, axis of symmetry is at x = 0.
Thus, focus is at (0, 2)
So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm
The distance of the light source from the bottom of the lamp is 2 cm.
The given parameters;
the top right edge of the lamp has a coordinate of (12,18)Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;
[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]
Thus, the distance of the light source from the bottom of the lamp is 2 cm.
"Your question is not complete, it seems to be missing the following information";
the top right edge of the lamp has a coordinate of (12,18)
Learn more here:https://brainly.com/question/14459938
In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in s) of the material
Answer:
2113 secondsExplanation:
The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;
[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;
[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16
[tex]\lambda[/tex] is the decay constant
t is the time taken for decay to occur = 8,450s
Before we can find the half life of the material, we need to get the decay constant first.
Substituting the given values into the formula above, we will have;
[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]
Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]
[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]
[tex]t_{1/2} = 2,112.8 secs[/tex]
Hence, the half life of the material is approximately 2113 seconds
g Calculate the maximum wavelength of light that will cause the photoelectric effect for potassium. Potassium has work function 2.29 eV = 3.67 x 10–19 J.
Answer:
λ = 5.4196 10⁻⁷m, λ = 541.96 nm this is green ligh
Explanation:
The photoelectric effect was explained by Eintein assuming that the light was made up of particles called photons and these collided with the electrons taking them out of the material.
K = h f -Ф
where K is the kinetic energy of the ejected electrons, hf is the energy of the light quanta and fi is the work function of the material.
The speed of light is related to wavelength and frequency
c = λ / f
f = c /λ
we substitute
K = h c / λ - Φ
for the case that they ask us the kinetic energy of the electons is zero (K = 0)
h c / λ = Ф
λ = h c / Ф
we calculate
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.67 10⁻¹⁸
λ = 5.4196 10⁻⁷m
let's take nm
lam = 541.96 nm
this is green light
An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
The distance from the Sun to Earth is approximately 149,600,000 km. The distance from the Sun to Venus is approximately 108,200,000 km. The elongation angle αα is the angle formed between the line of sight from Earth to the Sun and the line of sight from Earth to Venus. Suppose that the elongation angle for Venus is 10∘.10 ∘. Use this information to find the possible distances between Earth and Venus.
Answer:
335206922km
Explanation:
Pls see attached file
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of the circular motion of the alpha particles is closest to: A. 0.25 μs B. 0.16 μs C. 0.49 μs D. 0.40 μs E. 0.33 μs
Answer:
Option B: T ≈ 0.16 μs
Explanation:
We are given;
Mass; m = 6.68 × 10^(-27) kg
Magnetic field;B = 0.80 T
Charge;q = 2e
Now, e is the charge on an electron and it has a value of 1.6 × 10^(-19) C
So, q = 2 × 1.6 × 10^(-19)
q = 3.2 × 10^(-19) C
The period of the circular motion of the alpha particles moving along a in the presence of the magnetic field is given by;
T = 2πm/qB
Where ;
m, q and B are as stated earlier.
Plugging in the relevant values, we have;
T = (2π × 6.68 × 10^(-27))/(3.2 × 10^(-19) × 0.8)
T = 0.16395 × 10^(-6) s
This can also be written as;
T ≈ 0.16 μs
An 1,820 W toaster, a 1,420 W electric frying pan, and a 55 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.)
Required:
a. What current is drawn by each device?
b. Will this combination blow the 15-A fuse?
Answer:
toaster- 15.1A
electric frying pan- 11.8 A
lamp- 0.5 A
b) The combination will blow the fuse.
Explanation:
When devices are connected in parallel, the potential difference across each of the devices is the same but the current through each is different. Hence;
V= 120 V
Power= IV
For the toaster;
I= 1820/120 = 15.1 A
For the electric frying pan;
I= 1420/120 = 11.8 A
For the lamp;
55/120 = 0.5 A
Total current = 15.1 +11.8 + 0.5 = 27.4 A
The combination will blow the fuse.
Explanation:
step one:
Given data
power of toaster= 1,820 W
power of electric frying pan= 1,420 W
power of lamp= 55 W
current of the outlet= 15 A
voltage of outlet = 120 V
step two
since all three appliances are connected in parallel to the socket outlet, they will use the same voltage of 120 V and the currents will be different across each appliance,
Hence the current across the Toaster will be I₁
using P=I₁V we have
I₁= P/V
I₁= 1820/120 = 15.16 A
A. The current drawn by each device
the current across the electric frying pan will be I₂
using P=I₂V we have
I₂= P/V
I₂= 1420/120 = 11.83 A
the current across the lamp will be I₃
using P=I₃V we have
I₃= P/V
I₃= 55/120 = 0.45 A
therefore the total current drawn by all appliances will be
Total current = I₁+I₂+I₃= 15.16 +11.83+ 0.45= 27.44
B. Will this combination blow the 15-A fuse?
27.44 A > 15 A by 45% ...and this will make fuse to blow
How does a negative ion differ from an uncharged atom of the same
element?
O A. The ion has a greater number of protons.
B. The ion has fewer protons.
O C. The ion has a greater number of electrons.
O D. The ion has fewer neutrons.
Answer:
C if it is a negitive ion it has more electrons because protons determine what element it is
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the fundamental frequency
Answer:
f₀ = 158.12 HertzExplanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
[tex]V = \sqrt{\frac{T}{\mu} }[/tex] where T is the tension in the string and [tex]\mu[/tex] is the density of the string
Given T = 600N and [tex]\mu[/tex] = 0.015 g/cm = 0.0015kg/m
[tex]V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s[/tex]
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
A step-down transformer is used for recharging the batteries of portable devices. The turns ratio N2/N1 for a particular transformer used in a CD player is 2:29. When used with 120-V (rms) household service, the transformer draws an rms current of 180 mA.
Find the rms output voltage of the transformer
Answer:
8.28 V
Explanation:
Using,
N2/N1 = V2/V1.................. Equation 1
Where N2/N1 = Turn ratio of the transformer, V1 = primary/input voltage, V2 = output/secondary voltage
make V2 the subject of the equation
V2 = (N2/N1)V1............ Equation 2
Given: N2/N1 = 2:29 = 2/29, V1 = 120 V
Substitute these values into equation 2
V2 = (2/29)120
V2 = 8.28 V
Hence the rms output voltage of the transformer = 8.28 V
zeugen and yardang differences
Answer:
Yardangs are formed on vertical strata while zeugen on horizontal strata. ... Yardangs are formed on vertical hard/soft layers of rock, while zeugen (this is its plural form) are formed on horizontal bands of hard/soft rocks giving it a more mushroom-like shape. The Great Sphinx of Giza has been sculpted in a yardang
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a speed of 1.5 m/s. The coefficient of kinetic friction between the box and the incline is 0.34. How far does the box slide down the incline before coming to rest?
Answer:
The box will cover a distance of 0.9199m before coming to rest
Explanation:
We are given;
Angle of tilt; θ = 12°
Speed of sliding down; u = 1.5 m/s
Coefficient of kinetic friction; μ = 0.34
We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.
Thus,
The components of the weight of the block would be;
Fx = mg sinθ = mg sin 12
Fy = mg cosθ = mg cos 12
For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;
Normal force; Fn = mg cos 12
Now formula for the frictional force would be given by;
Ff = μmg cos 12
So, Ff = 0.34mg cos 12
So, the net force along the inclined plane is;
Fnet = Fx - Ff
Fnet = mg sin 12 - 0.34mg cos 12
Where Fnet = mass x acceleration.
Thus;
ma = mg sin 12 - 0.34mg cos 12
m will cancel out to give;
a = g sin 12 - 0.34g cos 12
a = 9.81(0.2079) - 0.34(9.81 × 0.9781)
a = -1.223 m/s²
According to Newton's equation of motion, we have;
(v² - u²) = 2as
s = (v² - u²)/2a
Final velocity is zero. Thus;
s = (0² - 1.5²)/(2 × -1.223)
s = -2.25/-2.446
s = 0.9199 m
Thus, the box will cover 0.9199m before coming to rest
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?
Answer:
Explanation:
Relation between flux and inductance is as follows
φ = Li
where φ is flux associated with induction of inductance L when a current i flows through it
putting the values
3.25 x 10⁻³ x 800 = L x 2.9
L = .9 H
for induced emf in an induction , the relation is
emf induced = L di / dt
Putting the values
7.5 x 10⁻³ = .9 x di / dt
di / dt = 8.33 x 10⁻³ A / s
(a) The self inductance of the solenoid is 0.897 H.
(b) The magnitude of the rate of change of the current is 0.00836 A/s.
The given parameters;
number of turns, N = 800 turnscurrent in the solenoid, I = 2.9 flux through the solenoid, Ф = 3.25 x 10⁻³ WbThe self inductance of the solenoid is calculated as follows;
[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]
The magnitude of the rate of change of the current is calculated as follows;
[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]
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Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
Determine the value of the current in the solenoid so that the magnetic field at the center of the loop is zero tesla. Justify your answer.
Answer:
I will explain the concept of magnetic field and how it can be calculated.
Explanation:
The formula for magnetic field at the center of a loop is given as
B = μ[tex]_{o}[/tex]I / 2R
where B is the magnetic field
R is the radius of the loop
I is the current
and μ[tex]_{o}[/tex] is the magnetic permeability of free space which is a constant 4π × [tex]10^{-7}[/tex] newtons/ampere²
If the magnetic field at the center of the loop is 0, then μ[tex]_{o}[/tex]I = 0
I = 0 which means there will be no current flow in the loop.
Specular reflection occurs where the light ray in the glass strikes the reflector. If no light is to enter the water, we require that there be reflection only. Which phenomenon prevents the light from entering the water?
Answer:
The critical angle phenomenon.
Explanation:
Critical angle in optics is the smallest angle of incidence of a wave, that will give total reflection of the wave. This phenomenon occurs at the boundary of two medium, where light will normally move from one medium to another.
To prevent light from entering the water, the angle of incidence of the light incident on the water must exceed the critical angle.
A polarized laser beam of intensity 285 W/m2 shines on an ideal polarizer. The angle between the polarization direction of the laser beam and the polarizing axis of the polarizer is 16.0 ∘. What is the intensity of the light that emerges from the polarizer?
Answer:
The intensity is [tex]I_1 = 263.35 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the beam is [tex]I = 285\ W/m^2[/tex]
The angle is [tex]\theta = 16^o[/tex]
The intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as
[tex]I_1 = I * cos^2 (\theta )[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
[tex]I_1 = 263.35 \ W/m^2[/tex]
Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.
Answer:
λ = 428.6 nm
Explanation:
Hello,
In this case, we must remember that the Young's double slit experiment is described by the expression :
d sin θ = m λ
For constructive interference , and:
d sin θ = (m + ½) λ
For destructive interference , whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference . Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation to compute the wavelength as shown below:
λ = 3x10⁻⁶ sin (30) / (3 +1/2)
λ = 4.286 10⁻⁷ m
Or in manometers:
λ = 428.6 nm
Best regards.
Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film.
Answer:
if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º
Explanation:
When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.
When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.
Also, when light penetrates the medium, it modifies its wavelength
λ = λ₀ / n
We take these two aspects into account, the condition for contributory interference is
d sin θ = (m + 1/2) λ
for destructive interference we have
d sin θ = m λ
in general this phenomenon is observed at 90º
2 d = (m +1/2) λ° / n
2nd = (m + ½) λ₀
A spark is generated in an automobile spark plug when there is an electric potential of 3000 V across the electrode gap. If 60 W of power is generated in a single spark that delivers a total charge of 3 nC, how long does it take for the spark to travel across the gap?
A. 50 ns
B. 75 ns
C. 125 ns
D. 150 ns
E. 225 ns 5
Answer:
The correct option is d
Explanation:
From the question we are told that
The electric potential is [tex]V = 3000 \ V[/tex]
The power is [tex]P = 60 \ W[/tex]
The charge delivered is [tex]q = 3nC = 3.0 *10^{-9} \ C[/tex]
Generally the power generated is mathematically represented as
[tex]P = I V[/tex]
=> [tex]I = \frac{P}{V }[/tex]
=> [tex]I = \frac{60 }{3000 }[/tex]
=> [tex]I = 0.02 \ A[/tex]
This current flow is mathematically represented as
[tex]I = \frac{q -q_o}{\Delta t }[/tex]
Where [tex]q_o[/tex] is the charge delivered at t=0 s which is 0s
So
[tex]0.02 = \frac{ (3.0 *10^{-9}) -0 }{t - 0 }[/tex]
[tex]t = 1.50 *10^{-7 } \ s[/tex]
[tex]t = 150 *10^{-9 } \ s[/tex]
A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?
a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s