Answer:
gshshs
Explanation:
hshsksksksbsbbshd
The "Pressure" meter allows you to read the pressure at different depths in the fluid. Place the pressure meter close to the bottom of the pool, and read the pressure. Slowly move the pressure meter toward the surface of the water in the pool and read the pressure at different depths in the pool. What happens to pressure in the fluid as the depth of the fluid decreases?
Answer:
The pressure near the surface of the pool will be less as compared that the bottom of the pool as water has weight. This is in relation to gravity
Explanation:
There is a relationship between volume and pressure. The increase in depth leads to an increase in volume and an increase in the force of gravity near the surface as compared to lifting and rising light pressure as light air rises and heavy air sinks.In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?
a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time
Answer:
b. The heavier one reaches the bottom first.
Answer:
B
Explanation:
The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.
Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 21.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s
Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]
[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]
(b) The energy stored in the inductor when 21 A current ;
[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]
In a single-slit diffraction pattern, the central fringe is 360 times as wide as the slit. The screen is 14,000 times farther from the slit than the slit is wide. What is the ratio /W, where is the wavelength of the light shining through the slit and W is the width of the slit
Answer:
0.01286
Explanation:
In a given single-slit, the central fringe (Y) is 360 times as wide as the slit (a). Then
2Y₁ = 360a
Y₁ = 360a/2
= 180a
The distance D = 14000a
In a given single-slit diffraction, the ratio = [tex]\dfrac{\lambda }{W}[/tex]
and since the angle is infinitesimally small;
sin θ ≅ tan θ = [tex]\dfrac{Y}{D}[/tex]
∴
For the first dark fringe;
Suppose: [tex]\dfrac{a}{2}sin \theta = \dfrac{\lambda }{2}[/tex]
then,
[tex]\dfrac{a}{2} \ \dfrac{Y_1}{D} = \dfrac{\lambda }{2}[/tex]
[tex]aY_1 = \lambda D[/tex]
[tex]\dfrac{\lambda }{a} = \dfrac{Y_1}{D}\\ \\ \\ \implies \dfrac{180\ a}{14000 \ a} \\ \\ \mathbf{\dfrac{\lambda }{a} = 0.01286}[/tex]
An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm
Answer:
960 nm
Explanation:
Given that:
wavelength = 640 nm
For the second (2nd) dark spot; the order of interference m = 1
Thus, the path length difference is expressed by the formula:
[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]
[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]
[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]
dsinθ = 960 nm
two electrons are separated by 1.10m, What is the magnitude of the electric force each electron exerts on the other?
Answer:
4.56×10¯⁷¹ N
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 1.10 m
Force (F) =?
NOTE:
Gravitational constant (G) = 6.67×10¯¹¹ Nm² /Kg²
Mass of electron = 9.1×10¯³¹ Kg
Mass of the two elections = M₁ = M₂ = 9.1×10¯³¹ Kg
Thus, we can obtain the force of attraction between the two elections as illustrated below:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × (9.1×10¯³¹)² / (1.1)²
F = 4.56×10¯⁷¹ N
Thus, the force of attraction between the two elections is 4.56×10¯⁷¹ N
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
Describe how the words Science and optics would appear when viewed in a plane mirror?
Answer:
Lateral inversion will occur in a plane mirror.
Explanation:
When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.
A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
What is the name of the compound br8P4
Answer:
Octabromine tetraphosphide
Explanation:
This compound has in its formula:
- Eight bromines
- Four phosphorous
8 → octa prefix
4 → tetra prefix
Right answer is Octabromine tetraphosphide
if a projectile travels in the air for 6 seconds when does the projectile reach its highest point
This question deals with projectile motion, which is a motion on both the x-axis and y-axis, simultaneously. The total time of flight of the projectile trajectory is given, while the time to reach the highest point of the projectile is required to be found.
The projectile will reach the highest point in "3 seconds".
The total time of flight of a projectile is the time during which the projectile remains in the air. For a projectile motion that ends up on the same horizontal level, from where it started, the time to reach the highest point, is equal to half of the total time of flight.
In other words, the projectile motion takes the same time, to go from the starting level to the highest point (i.e upward motion), as the time taken to reach the starting level from the highest point (i.e downward motion).
[tex]t = \frac{1}{2}T[/tex]
where,
t = time to reach the highest point = ?
T = total time of flight = 6 seconds
Therefore,
[tex]t - \frac{1}{2}(6\ seconds)[/tex]
t = 3 seconds
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A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.
What is the equation of motion?The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.
Given, the initial velocity of the projectile, u = 75 m/s
The time taken by the projectile, t = 3 sec
The acceleration due to gravity upward, g = - 9.8 m/s²
From the first equation of motion we can calculate the final velocity of the projectile:
v = u + at
v = u - gt
v = 75 - 9.8 ×(3)
v = 75 - 29.4
v = 45.6 m/s
Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.
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Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest
i. [tex]T = 36.8\:\text{N}[/tex]
ii. [tex]a = 2.45\:\text{m/s}^2[/tex]
iii. [tex]x = 1.23\:\text{m}[/tex]
Explanation:
Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]
Forces acting on m1:
[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]
Forces acting on m2:
[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]
Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us
[tex]m_2g - m_1g = m_2a + m_1a[/tex]
or
[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]
Solving for a,
[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]
[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]
We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).
[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]
Assuming that the two objects start from rest, the distance that they travel after one second is given by
[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]
Give reason why think before you use a simple cell ?
The disadvantages of simple cell are: It is not rechargeable. The battery needs to be disposed of after all the power has been used up. It can't produce electricity anymore. That is why, why think before you use a simple cell.
What are the benefits and drawbacks of simple cell?A battery designed to be used only once is called a simple cell. Small gadgets used in the house are frequently powered by simple cells.
The benefits of a simple cell include:
A simple cell can be used to power small electronic devices because of its modest size. (Games, lightsabers, radios on the go, cameras, hearing aids)Simple cell electrolyte is not very detrimental to the environment.Simple cells are reasonably priced.Among the drawbacks of a simple cell are:
The biggest drawback of a simple cell is that once it runs out of electricity, it cannot be replenished.Learn more about cell here:
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A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave
Answer:
a) [tex]v=206.896m/s[/tex]
b) [tex]L=6.749m[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=46Hz[/tex]
Length [tex]l=579m[/tex]
Total Mass [tex]T=4.3kg[/tex]
Tension [tex]T=3423[/tex]
a)
Generally the equation for velocity is mathematically given by
[tex]v=\sqrt{\frac{T}{\rho}}[/tex]
Where
[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]
Therefore
[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]
[tex]v=206.896m/s[/tex]
b)
Generally the equation for length of string is mathematically given by
[tex]L=\frac{3\lambda}{2}[/tex]
Where
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{206.89}{46}[/tex]
[tex]\lambda=4.498[/tex]
Therefore
[tex]L=\frac{3*4.498}{2}[/tex]
[tex]L=6.749m[/tex]
find the upward force in Newton when each of these is under water(density of 1g/cm3) a lump of iron of volume 2000cm3
Answer:
Upthrust = 19.6 N
Explanation:
When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.
Upthrust = density of liquid x gravitational force x volume of object
i.e U = ρ x g x vol
Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]
So that;
U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])
= 19.6 Kg m/[tex]s^{2}[/tex]
U = 19.6 Newtons
The upthrust on the iron is 19.6 N.
CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?
Answer:
θ' = 14.44 × [tex]10^{6}[/tex]
Explanation:
given data
total distance is d = 4820
radius = 1.4 cm
solution
we get here total angle by which the wheel rotates traveling is express as
⇒ [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex] ................1
and
total angle (θ) and the total distance (d) express as
⇒ d = r × θ ...............2
here r is radius
and here rotated through some other angle θ' so put value in given equation and find revolutions
⇒ d = (r+r)θ' ........3
here r = d/θ
so
⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]
so put value and get θ'
⇒ θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]
⇒ θ' = 14.44 × [tex]10^{6}[/tex] revwhat is threshold frequency?
Answer:
"the minimum frequency of radiation that will produce a photoelectric effect."
Explanation:
That answer was derived from gogle cuz my explanations was harder to explain but good luck
A string has its 4th harmonic at 31.5 Hz. What is the frequency of its third harmonic?
Answer:
The answer would be 7.5 Hz.
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].
Answer:
89.87m/s
Explanation:
Given the velocity function
v(t)=−t²+3t−2
In order to get the displacement function, we will integrate the velocity function as shown:
[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]
at t = 5
[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]
at t = -2
[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]
Required displacement = d(5) - d(-2)
Required displacement = 89.2 - (-0.67)
Required displacement = 89.2 + 0.67
Required displacement = 89.87m/s
A particle moves along X-axis in such a way that X-coordinate varies with time according to expression x= 2-5t+6t2 meters, Calculate the initial velocity of the particle?
A 5
v= dt/ dx =−5+12t
Initial velocity means at t=0, which is −5+0=−5.
Thus, −v=5n
The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
Your physics TA has a far point of 0.759 m from her eyes and is able to see distant objects in focus when wearing glasses with a refractive power of −1.35 D. Determine the distance between her glasses and eyes.
Answer:
[tex]d=0.019m[/tex]
Explanation:
From the question we are told that:
Far point [tex]x=0.759m[/tex]
Refractive power [tex]P=-1.35 D.[/tex]
Generally, the equation for Focal length is mathematically given by
[tex]F=\frac{1}{P}[/tex]
[tex]F=\frac{1}{-1.35}[/tex]
[tex]F=-0.74m[/tex]
Therefore
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
Where
[tex]u=o[/tex]
[tex]\frac{1}{-0.74}=\frac{1}{0}+\frac{1}{v}[/tex]
[tex]v=-0.74m[/tex]
Therefore,The between her glasses and eyes
[tex]d=x-v[/tex]
[tex]d=0.759-0.74m[/tex]
[tex]d=0.019m[/tex]
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
3. Some guitarists like the feel of a set of strings that all have the same tension. For such a guitar, the G string (196 Hz) has a mass density of 0.31 g/m. What is the mass density of the A string (110 Hz)
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
A light spectrum is formed on the screen using a diffraction grating. The entire apparatus made up of laser, grating and the screen is now immersed in a liquid with refractive index 1.33. Do the bright spots on the screen get closer together, farther apart, remain the same or disappear entirely? Explain
Answer:
the points are closer to each other
Explanation:
The expression for the diffraction of a grating is
d sin θ = m λ
sin θ = m λ / d (1)
where d is the distance between slits and m is the order of diffraction, the most general is to work in the order m = 1, the angle te is the angle of diffraction
When we immerse the apparatus in a medium with refractive index n = 1.33, the light emitted by the laser must comply
v = λ f
where v is the speed of light in the medium, the frequency remains constant
velocity and refractive index are related
n = c / v
v = c / n
we substitute
c / n = λf
λ = [tex]\frac{c}{f} \ \frac{1}{n}[/tex]
λ = λ₀ / m
where λ₀ is the wavelength in vacuum
we substitute is equation 1
d sin θ = m λ₀ / n
sin θ = λ₀/ n d
sin θ = [tex]\frac{1}{n}[/tex] sin θ₀
we can see that the value of the sine is redueced since the refractive index is greater than 1,
consequently the points are closer to each other
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
