Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
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Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
Which has more mass electron or ion?
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe
Answer:
0.0127m
Explanation:
Using
Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?
Answer:
The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Explanation:
We are told that the particle moves back and forth along a straight line with velocity v(t).
Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.
Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air
Answer:
The critical angle is [tex]i = 41.84 ^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the sugar solution is [tex]n_s = 1.5[/tex]
The index of refraction of air is [tex]n_a = 1[/tex]
Generally from Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]
Note that the angle of incidence in this case is equal to the critical angle
Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]
So
[tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]
[tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]
[tex]i = 41.84 ^o[/tex]