A converging lens 7.50 cm in diameter has a focal length of 330 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of resolving power of the human eye. Part A If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.10 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 600 nm
Answers
Answer:
D Is 430m
Explanation:
See attached file
Related Questions
A father and his son want to play on a seesaw. Where on the seesaw should each of them sit to balance the torque?
Answers
Answer:
A The father should sit closer to the pivot.
C The longer wrench makes the job easier because less force is needed when there is more distance from the pivot.
A As far from the head of the hammer as possible because this will maximize torque.
D at the opposite side of the seesaw towards the middle
:) gl
Explanation:
If a father and his son want to play on a seesaw then to balance the torque of the seesaw the father should sit near the pivot as he had more weight as compared to his son, while the son should sit a little farther from the pivot point as compared to his father.
What is the mechanical advantage?
Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.
Mechanical advantage = output force(load) /input force (effort)
As given in the problem statement If a father and his son wish to play on a seesaw,
The father should sit close to the pivot because he weighs more than his son, and the son should sit a little farther away from the pivot point than his father. This will help balance the torque of the seesaw.
Thus, the father should sit near the pivot on the one side and the son should sit a little farther from the pivot of a seesaw on the other side.
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"If a beam of monochromatic light is passed though a slit of width 15 μm and the second order dark fringe of the diffraction pattern is at an angle of 5.2o from the central axis, what is the wavelength of the light?"
Answers
Answer:
λ= 5.4379 10⁻⁷ m = 543.79 nm
Explanation:
The phenomenon of diffraction is described by the expression for destructive diffraction is
a sin θ = (m + 1/2) λ
λ = a sin θ / (m + 1/2)
let's reduce the magnitudes to the SI system
a = 15 um = 15 10⁻⁶ m
m = 2
θ = 5.2º
Let's calculate
λ = 15 10⁻⁶ sin 5.2 / (2 +1/2)
λ = 5.4379 10⁻⁷ m
Let's reduce to nm
λ= 5.4379 10⁻⁷ m = 543.79 nm
a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density
Answers
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
Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.
For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:
Rate of energy loss = AεσT4
where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:
σ = 5.67 x 10-8 J/(s m2 K4)
Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.
a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts
b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC
c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g
Answers
Answer:
a) 360.7 J/s
b) 16.23 °C
c) 34.48 g
Explanation:
The mass of the person = 80 kg
The person is a perfect emitter, ε = 1
surface area of the person = 2.5 m^2
a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K
and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K
Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )
where σ = 5.67 x 10^-8 J/(s m^2 K^4)
substituting values, we have
E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s
b) If they have specific heat about equal to that of water = 1 Cal/kg-°C
but 1 Cal = 1 kcal = 10^3 cal
specific heat of person is therefore = 10^3 cal/kg-°C
heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr
heat lost in 1 hour = 1 x 1298520 = 1298520 J
This heat lost = mcΔT
where ΔT is the temperature fall
m is the mass
c is the specific heat equivalent to that of water
the specific heat is then = 10^3 cal/kg-°C
equating, we have
1298520 = 80 x 10^3 x ΔT
1298520 = 80000ΔT
ΔT = 1298520/80000 = 16.23 °C
c) 1298520 J = 1298520/4184 = 310.35 Cal
density of fat = 9 Cal/g
gram of fat = 310.35/9 = 34.48 g
What is the answer?
Answers
Answer: i think it is d. none of them.
Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answers
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
A non-ideal battery has a 6.0-V emf and an internal resistance of 0.6 l. Determine the terminal voltage (in volts) when the current drawn from the battery is 1.0 A
Answers
=6,0 --0.6 =5.4 volts terminal volts 5.4 volts
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answers
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
Several books are placed on a table. These books have a combined weight of 25 N and cover an area of 0.05 m2. How much pressure do the books exert on the table? The pressure the books apply to the table top is __ Pa.
Answers
Answer:
500 PascalsExplanation:
[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]
A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) about 20 times per second.part a. Determine the number of photons in one 80-mJ pulse.part b. Determine the average power of photons in one 80-mJ pulse during 1 s.
Answers
Answer:
a
[tex]n = 1.119 *10^{18} \ photons[/tex]
b
[tex]P = 1.6 \ W[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 2780 nm = 2780 *10^{-9} \ m[/tex]
The energy is [tex]E = 80 mJ = 80 *10^{-3} \ J[/tex]
This energy is mathematically represented as
[tex]E = \frac{n * h * c }{\lambda }[/tex]
Where c is the speed of light with a value [tex]c = 3.0 *10^{8} \ m/s[/tex]
h is the Planck's constant with the value [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
n is the number of pulses
So
[tex]n = \frac{E * \lambda }{h * c }[/tex]
substituting values
[tex]n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }[/tex]
[tex]n = 1.119 *10^{18} \ photons[/tex]
Given that the pulses where emitted 20 times in one second then the period of the pulse is
[tex]T = \frac{1}{20}[/tex]
[tex]T = 0.05 \ s[/tex]
Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as
[tex]P = \frac{E}{T}[/tex]
substituting values
[tex]P = \frac{ 80 *10^{-3}}{0.05}[/tex]
[tex]P = 1.6 \ W[/tex]
how many stars are in our solar system?
Answers
Answer:
there are over 100 billion stars in our galaxy.
A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What minimum energy is required to change the orbit to one for which the height of the satellite is 3R above the surface of the planet
Answers
Answer:
ΔE = GMm/24R
Explanation:
centripetal acceleration a = V^2 / R = 2T/mr
T= kinetic energy
m= mass of satellite, r= radius of earth
= gravitational acceleration = GM / r^2
Now, solving for the kinetic energy:
T = GMm / 2r = -1/2 U,
where U is the potential energy
So the total energy is:
E = T+U = -GMm / 2r
Now we want to find the energy difference as r goes from one orbital radius to another:
ΔE = GMm/2 (1/R_1 - 1/R_2)
So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R
ΔE = GMm/2R (1/3 - 1/4)
ΔE = GMm/24R
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answers
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answers
Answer:
E = 1336.71875 J
Explanation:
We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF
Capacitor 2; C2 = 23.0 µF
Capacitor 3; C3 = 29.3 µF
Supply voltage;V = 125 V
Formula for capacitance in series is;
Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......
Thus, equivalent capacitance is;
C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F
Now, the formula for maximum energy stored is;
E = ½ × C(eq) × V²
E = ½ × 0.1711 × 10^(-6) × 125²
E = 1336.71875 J
A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or occurring at constant pressure. Rank those three processes in order of the work required of an external agent, least to greatest. A : adiabatic, isothermal, constant pressure B : constant pressure, isothermal, adiabatic C : adiabatic, constant pressure, isothermal D : isothermal, adiabatic, constant pressure E : constant pressure, adiabatic, isothermal
Answers
Answer:
B. constant pressure, isothermal, adiabatic
Explanation:
A quantity of an ideal gas is compressed to half its initial volume.
The process may be adiabatic, isothermal or occurring at constant pressure.
Adiabatic-constant heat
Constant pressure or isobaric
Isothermal or constant temperature
An external agent is a system that does work on a system or a machine.
This external agent applies force , or changes the state of the body it is acting on.
In order of the work required of an external agent, least to greatest
The following processes will be arranged.
constant pressure, isothermal, adiabatic
A) A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied? (1 point)
1 m
4 m
0.08 m
2 m
Answers
Answer:
Option (A) : 1m
Explanation:
According to Hooke's law:
F (spring elastic force) =
k( spring const.) * x(displacement)
Case-1
2 N = k * 0.4m
k = 5
Case- 2
5 N = 5 * x
x ( displacement) = 1 m
The displacement of the spring if a 5-N force was applied is equal to 1m. Therefore, option (1) is correct.
What is Hooke's law?The strain and stress are proportional to each other, and this is called Hooke’s Law. Hooke’s law states that the strain is proportional to the stress applied within the elastic limit of the material.
When the materials are stretched, the atoms or molecules deform and when the stress is removed, they will return to their original state.
The mathematical equation for Hooke's law is as follows:
F = –kx
where F is the force, x is displacement, and k is the spring constant in N/m.
Given, F = 2N and x = 0.4m
F = -kx
2 N = - k (0.4m)
k = 5 N/m where the negative sign is omitted.
Now, the spring constant of the spring, k = 5 N/m and F = 5N
F = -kx
5 N = - (5 N/m)(x)
x = - 1m
Therefore, the displacement of the spring is 1 m.
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As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answers
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answers
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the first-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm
Answers
Answer:
[tex]\theta_1 = 0.400^o[/tex]
[tex]\theta_2 =0.378^o[/tex]
Explanation:
From the question we are told that
The number of slits per cm is k = [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]
The order of the maxima is n = 1
The wavelength are [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]
The spacing between the slit is mathematically represented as
[tex]d = \frac{ 0.01}{k}[/tex]
=> [tex]d = \frac{ 0.01}{161}[/tex]
=> [tex]d = 6.211 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]n\lambda = d \ sin \theta[/tex]
At [tex]\lambda_1[/tex]
[tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_1 = 0.400^o[/tex]
At [tex]\lambda_2[/tex]
[tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_2 =0.378^o[/tex]
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answers
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
(4) Use the preliminary observations to answer these questions; Compared to no polarizer or analyzer in the optical path, by what percent does the light intensity decrease when (a) The polarizer is introduced into the optical path? (b) The both polarizer and analyzer are introduced into the optical path?
Answers
Answer:
a) I = I₀/2, b) I = I₀/2 cos² θ
Explanation:
To answer these questions, let's analyze a little the way of working of a polarized
* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is
i = I₀ / 2
and is polarized in the direction of the polarizer
* when a polarized light reaches the analyzer it must comply with Malus's law
I = I₁ cos² θ
where the angle is between the polarized light and the analyzer.
With this, let's answer the questions
a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of
I = I₀/2
b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is
I = I₀/2 cos² θ
where the final value depends on the angle between the polarizer and the analyzer.
Let's look at two extreme cases
θ = 0 I = Io / 2
θ = 90º I = 0
A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory?
Answers
The bullet's vertical velocity at time [tex]t[/tex] is
[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
At its highest point, the bullet's vertical velocity is 0, which happens
[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]
(or about 140 s, if you're keeping track of significant figures) after being fired.
what happened when aniline is treated with benzene diazonium chloride
Answers
Answer:
p-aminoazobenzene is formed
Explanation:
The reaction of benzene diazonium chloride and aniline takes place in a basic medium and leads to the formation of an azo compound which is also a dye. The terminal diazonium nitrogen of the benzene diazonium ion is coupled to the aniline at the para-position. The product of the reaction, p-aminoazobenzene is a yellow dye.
Benzene diazonium chloride is prepared by diazotization of aniline in the presence of hydrochloric acid. The full reaction of aniline and benzene diazonium chloride is shown in the image attached to this answer.
Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail (in m) utilizing 495 MHz radar?
Answers
Answer:
0.61 m
Explanation:
The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.
using the relationship
c = fλ
where
c is the speed of light in vacuum = 3 x 10^8 m/s
f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz
λ is the wavelength = ?
λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m
answer to your question is 0.6m
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answers
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under the influence of the force field. F(x, y, z)= z^2i + 4xyj + 5y^2kFind the work done.
Answers
Answer:
0 J
Explanation:
Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk
F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz
W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z
We now evaluate the work done for the different regions
W₁ = work done from (0,0,0) to (1,0,0)
W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J
W₂ = work done from (1,0,0) to (1,5,1)
W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J
W₃ = work done from (1,5,1) to (0,5,1)
W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J
W₄ = work done from (0,5,1) to (0,0,0)
W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J
The total work done W is thus
W = W₁ + W₂ + W₃ + W₄
W = 0 J + 176 J - 51 J - 125 J
W = 176 J - 176 J
W = 0 J
The total work done equals 0 J
A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples does it produce when a 3.25 m focal length eyepiece is used? ✕
Answers
Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
Unpolarized light is passed through three successive Polaroid filters, each with its transmission axis at 45.0° to the preceding filter. What percentage of light gets through?
Answers
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
What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.
Answers
Answer:
C
They allow light to pass only if their directions of polarizations are exactly 90° apart.
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?
Answers
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
