Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?
Answer:
L = 0.8 m
Explanation:
Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:
Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m
but,
Δx = λL/d
λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m
d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m
L = Distance between slits and screen = ?
Therefore, using the values, we get:
2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)
L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)
L = 0.8 m
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate
Answer:
The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.
Explanation:
Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.
a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density
Answer:
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
Explanation:
given data
frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz
we consider here intensity I = 0.2 W/m²
solution
we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s
and take energy density is E
so here
E × C = I
E = [tex]\frac{I}{C}[/tex] ................1
here C = 3 × [tex]10^{8}[/tex] m/s
so photon density is
photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex] ...............2
photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
What is the difference between matter and energy
Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
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Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?
Explanation:
Given that,
Linear speed of both disks is 5 m/s
Mass of disk 1 is 10 kg
Radius of disk 1 is 35 cm or 0.35 m
Mass of disk 2 is 3 kg
Radius of disk 2 is 7 cm or 0.07 m
(a) The angular velocity of disk 1 is :
[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]
(b) The angular velocity of disk 2 is :
[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]
(c) The moment of inertia for the two disk system is given by :
[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]
Hence, this is the required solution.
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
What is the direction of the net gravitational force on the mass at the origin due to the other two masses?
Answer:
genus yds it's the
Explanation:
xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
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Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)
d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)
= 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have
Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex] (1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex] (2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?
Answer:
The induced current is clockwise
Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m
Answer:
0.03/π m/min
Explanation:
See attached file pls
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor through paddles placed on the skin, causing charge to flow through the heart. Assume that the capacitor is originally charged with 5.0 kV .Part AWhat is the charge initially stored on the capacitor?3×10−9 C7.5×104 C7.5×10−2 C7.5×10−5 CPart BWhat is the energy stored on the capacitor?What is the energy stored on the capacitor?1.9×108 J380 J190 J1.9×10−4 JPart CIf the resistance between the two paddles is 100 Ω when the paddles are placed on the skin of the patient, how much current ideally flows through the patient when the capacitor starts to discharge?5×105 A50 A2×10−2 A5×10−2 APart DIf a defibrillator passes 17 A of current through a person in 90 μs . During this time, how much charge moves through the patient?If a defibrillator passes 17 {\rm A} of current through a person in 90 {\rm \mu s} . During this time, how much charge moves through the patient?190 mC1.5 C1.5 mC17 C
Answer:
a) q = 7.5 10⁻² C , b) 190 J , c) I₀ = 50 A , d) 1.5 mC
Explanation:
The expression for capacitance is
C = q / DV
q = C DV
let's reduce the magnitudes to the SI system
ΔV = 5 kV = 5000 V
C = 15 μF = 15 10⁻⁶ F
t = 90 μs = 90 10⁻⁶ s
q = 15 10⁻⁶ 5000
q = 7.5 10⁻² C
b) the energy in a capacitor is
U = ½ C ΔV²
U = ½ 15 10⁻⁶ 5000²
U = 1,875 10² J
answer 190 J
c) At the moment the discharge begins, all the current is available and it decreases with time,
whereby
V = I R
in the first instant I = Io
I₀ = V / R
I₀ = 5000/100
I₀ = 50 A
but this is for a very short time
answer 50 A
d) The definition of current is
i = dq / dt
in this case they give us the total current and the total time, so we can find the total charge
i = q / t
q = i t
q = 17 90 10⁻⁶
q = 1.53 10⁻³ C
answer is 1.5 mC
In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
Krishna and Seldon now try a homework problem. A policeman sitting in his unmarked police car sees an approaching motorcyclist go through a red light two blocks away. He turns on his siren at a frequency of 1000 Hz as the motorcyclist heads directly toward him at 61 mph (27.27 m/s). What frequency does the motorcyclist hear? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz What frequency does the motorcyclist hear when stopped with the police car approaching at 61 mph (27.27 m/s)? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz
Answer:
Explanation:
We shall apply formula of Doppler's effect
Here source is fixed and observer is approaching the source
f = f₀ x [(V + v ) / V ]
f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .
f = 1000 x [(331 + 27.27 ) / 331 ]
= 1082 .4 Hz
This is the frequency heard by motorcyclist .
When police car is approaching him when he is stopped
f = f₀ x [V /(V - v ) ]
v is velocity of police car .
= 1000 x 331 / (331 - 27.27)
= 1090 Hz
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answer:
They both have the same acceleration
A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf
Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?
Answer:
Δy = 1 10⁻⁴ m
Explanation:
In double-slit experiments the constructive interference pattern is described by the equation
d sin θ = m λ
In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle
tan θ = y / L
as the angles are small,
tan θ = sin θ / cos θ = sin θ
substituting
sin θ = y / L
d y / L = m λ
y = m λ / d L
let's apply this formula for each wavelength
λ = 450 nm = 450 10⁻⁹ m
m = 5
d = 5.0 mm = 5.0 10⁻³ m
y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)
y₁ = 3.21 10⁻⁴ m
we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m
y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)
y₂= 4.21 10⁻⁴ m
the separation of these two lines is
Δy = y₂ - y₁
Δy = (4.21 - 3.21) 10⁻⁴ m
Δy = 1 10⁻⁴ m
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Learn more about the average density here:
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