Answer:
Explanation:
Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/
α = 26.56°
Laser angle with Normal = 90 - 26.56 = 63.44 °
Assuming a red laser, refractive index in water is 1.331.
Angle of refraction in water is given by:
Ref Ind = Sin i / Sin r
1.331 = Sin 63.44 / Sin r
Sin r = 0.8945 / 1.331 = 0.6721
Angle r = 42.22°
For the path in water:
Tan 42.22 = x / 3.2
x = 2.9m where x is the lateral displacement of the laser ince it hits the water
So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool
Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the universe. The universe is very big and very old. The universe consists of matter.
Answer:
Explanation:
That Universe Consists of Matter
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.
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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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
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
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
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/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 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].
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IMPORTANT ANSWER ALL 3 PLEASE!
Answer:
4. Liters
5. Celsius
6. Grams
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:
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]
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
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.
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]
Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?
Answer:
29.5°
Explanation:
To find the distance d
d = 1E10^-2/8500lines
= 1.17x 10-6m
But wavelength in first order maximum is 577nm
and M = 1
So
dsin theta= m. Wavelength
Theta= sin^-1 (m wavelength/d)
= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6
= 493*10^-3= sin^-1 0.493
Theta = 29.5°
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
Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
[tex]m1U1+ m2U2= m1V1+m2V2[/tex]
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
[tex]m1U1+ m2U2= V(m1+m2)[/tex]
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
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
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.
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]
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 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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Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct
Answer:
Both Ellen and Mary are correct.
Explanation:
Both are correct, it's just different ways of saying the same thing.
When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position
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!
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.
For this "type" of motion, displacement (Δx) can be determined by:
[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]
[tex]v_{i}[/tex] is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]
[tex]\Delta x = 30 - 2.36[/tex]
[tex]\Delta x[/tex] = - 42
Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.
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.
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:
