Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
It is just as difficult to accelerate a car on a level horizontal surface on the Moon as it is here on Earth because
Answer:
Mass of the car is independent of gravity
Explanation:
Here, we want to state the reason why even though we have the acceleration due to gravity absent on the moon, it is still difficult to accelerate a car on a level horizontal level on the moon.
The answer to this is that the mass of the car that we want to accelerate is independent of gravity.
Had it been that gravity has an effect on the mass of the said car, then we might conclude that it will not be difficult to accelerate the car on a horizontal surface on the moon.
But due to the fact that gravity has no effect on the mass of the car to be accelerated, then the problem we have on earth with accelerating the car is the same problem we will have on the moon if we try to accelerate the car on a horizontal level surface.
Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surface, fixed to the ground with a centre O and radius of 0.1m. The spheres are joined by the cord shown in length π/20 m; determine the angles α and β corresponding to the position of equilibrium of the spheres with respect to the vertical passing through O.
Answer:
α = π/3
β = π/6
Explanation:
Use arc length equation to find the sum of the angles.
s = rθ
π/20 m = (0.1 m) (α + β)
π/2 = α + β
Draw a free body diagram for each sphere. Both spheres have three forces acting on them:
Weight force mg pulling down,
Normal force N pushing perpendicular to the surface,
and tension force T pulling tangential to the surface.
Sum of forces on A in the tangential direction:
∑F = ma
T − m₁g sin α = 0
T = m₁g sin α
Sum of forces on B in the tangential direction:
∑F = ma
T − m₂g sin β = 0
T = m₂g sin β
Substituting:
m₁g sin α = m₂g sin β
m₁ sin α = m₂ sin β
(1 kg) sin α = (√3 kg) sin (π/2 − α)
1 sin α = √3 cos α
tan α = √3
α = π/3
β = π/6
You simultaneously shine two light beams, each of intensity I0, on an ideal polarizer. One beam is unpolarized, and the other beam is polarized at an angle of exactly 30.0∘ to the polarizing axis of the polarizer. Find the intensity of the light that emerges from the polarizer. Express your answer in term of I0 .
Answer:
The emerging intensity is equal to 0.75[tex]I_{o}[/tex]
Explanation:
The initial intensity of the light = [tex]I_{o}[/tex]
The angle of polarization β = 30°
We know that the polarized light intensity is related to the initial light intensity by
[tex]I[/tex] = [tex]I_{0} cos^{2}\beta[/tex]
where [tex]I[/tex] is the emerging polarized light intensity
inserting values gives
[tex]I[/tex] = [tex]I_{0} cos^{2}[/tex] 30°
[tex]cos^{2}[/tex] 30° = [tex](cos 30)^{2}[/tex] = [tex](\frac{\sqrt{3} }{2} )^{2}[/tex] = 0.75
[tex]I[/tex] = 0.75[tex]I_{o}[/tex]
4. A diver is 20 m underwater and they are startled by a shark. They are tempted to take a big breath of air, drop their gear, and swim to the surface while holding their breath. Explain why this is dangerous g
Answer:
Explanation:
The air enters their lungs at the same pressure as the water at that depth.
If they hold their breath as they rise to atmospheric pressure, the expanding volume of air (due to decreasing pressure) trapped in their lungs will hyperextend the alveoli in their lungs, likely tearing blood lines and risking death by drowning in their own blood.
Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.
Requried:
What is the frequency the horns emit?
Answer: f ≈ 8.5Hz
Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.
For a source moving and a stationary observer, to determine the frequency:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
where:
[tex]f_{0}[/tex] is frequency of observer;
[tex]f_{s}[/tex] is frequency of source;
c is the constant speed of sound c = 340m/s;
[tex]v_{s}[/tex] is velocity of source;
Rearraging for frequency of source:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]
Replacing and calculating:
[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]
[tex]f_{s} = 9.(0.9412)[/tex]
[tex]f_{s} =[/tex] 8.5
Frequency the horns emit is 8.5Hz.
A magnetic field near the floor points down and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counter-clockwise?
a. clockwiseb. counter-clockwise c. no curly E
Answer:
when a magnetic field near the floors points down and is increasing then the electric field curl (a) clockwise.
Explanation:
The magnetic field this is the area that is around a magnet which there is presence of magnetic force. The Moving electric charges can create magnetic fields. we say In physics, that the magnetic field is a field that passes through space and which makes a magnetic force move electric charges.
The Non-coulomb electric field curls ; ( B ) counterclockwise
Non-coulomb electric field also known as induced EMF is the Negative time rate of change of a magnetic flux in a closed loop through the loop. Non-coulomb electric field is expressed as ; Fnc = qEnc
Given that the magnetic field points downwards and the value of the electric field ( ε ) is increasing ( i.e. ε > 0 ) The direction of the non-coulomb electric field will curl in a counter-clockwise direction.
Hence we can conclude that The Non-coulomb electric field curls in a counterclockwise direction.
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5. An electron is moving north in a region where
the magnetic field is south. The magnetic force
exerted on the electron is:
A.Zero
B. up
C. down
D. east
E. west
Answer:
eastlove india
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.
Calculate di/dt
di/dt = _________.
Answer:
[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 450 \ turns[/tex]
The radius is [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]
The position from the center consider is x = 3.45 cm = 0.0345 m
The induced emf is [tex]e = 8.20 *10^{-6} \ V/m[/tex]
Generally according to Gauss law
[tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]
Where A is the cross-sectional area of the solenoid which is mathematically represented as
[tex]A = \pi r ^2[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]
=> [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
The value of di/dt from the given values of the solenoid electric field is;
di/dt = 7.415 A/s
We are given;
Number of turns; N = 450 per m
Radius; r = 1.17 cm = 0.0117 m
Electric Field; E = 8.2 × 10⁻⁶ V/m
Position of electric field; r' = 3.45 cm = 0.0345 m
According to Gauss's law of electric field;
∫| E*dl | = |-d∅/dt |
Now, ∅ = BA = μ₀niA
where;
n is number of turns
i is current
A is Area
μ₀ = 4π × 10⁻⁷ H/m
Thus;
E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)
Since we are looking for di/dt, then we have;
E(2πr') = (di/dt)(μ₀nA)
Making di/dt the subject of the formula gives;
di/dt = E(2πr')/(μ₀nA)
Plugging in the relevant values gives us;
di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)
di/dt = 7.415 A/s
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A nozzle with a radius of 0.22 cm is attached to a garden hose with a radius of 0.89 cm that is pointed straight up. The flow rate through hose and nozzle is 0.55 L/s.
Randomized Variables
rn = 0.22 cm
rh = 0.94 cm
Q = 0.55
1. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.
2. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the same flow rate.
Answer:
1. 0.2m
1. 66m
Explanation:
See attached file
The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;
1) The maximum height when the water leaves the hose is: Δy = 0.20 m
2) The maximum height of the water leaves the nozzle is: Δy = 68.6m
Given parameters
The flow rate Q = 0.55 L/s = 0.55 10⁻³ m³ / s Nozzle radius r₁ = 0.22 cm = 0.22 10⁻² m Hose radius r₂ = 0.94 cm = 0.94 10⁻² mTo find
1. Maximum height of water in hose
2. Maximum height of water at the nozzle
Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:
The continuity equation. It is an expression of the conservation of mass in fluids.
A₁v₁ = A₂.v₂
Bernoulli's equation. Establishes the relationship between work and the energy conservation in fluids.P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂
Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.
1, Let's find the exit velocity of the water in the hose.
Let's use subscript 1 for the nozzle and subscript 2 for the hose.
The continuity equation of the flow value that must be constant throughout the system.
Q = A₁ v₁
v₁ = [tex]\frac{Q}{A_1 }[/tex]
The area of a circle is:
A = π r²
Let's calculate the velocity in the hose.
A₁ = π (0.94 10⁻²) ²
A₁ = 2.78 10⁻⁴ m²
v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]
v₁ = 1.98 m / s
Let's use Bernoulli's equation.
When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2
½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂
y₂-y₁ = ½ [tex]\frac{v_i^2 - v_2^2}{g}[/tex]
At the highest point of the trajectory the velocity must be zero.
y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]
Let's calculate
y₂-y₁ = [tex]\frac{1.98^2}{2 \ 9.8}[/tex]
Δy = 0.2 m
2. Let's find the exit velocity of the water at the nozzle
A₁ = π r²
A₁ = π (0.22 10⁻²) ²
A₁ = 0.152 10⁻⁴ m / s
With the continuity and flow equation.
Q = A v
v₁ = [tex]\frac{Q}{A}[/tex]
v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]
v₁ = 36.67 m / s
Using Bernoulli's equation, where the speed of the water at the highest point is zero.
y₂- y₁ = [tex]\frac{v^1^2}{g}[/tex]
Let's calculate.
Δy = [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]
Δy = 68.6m
In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:
1) The maximum height when the water leaves the hose is:
Δy = 0.20 m
2) The maximum height of the water when it leaves the nozzle is:
Δy = 68.6 m
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If a vacuum pump reduces the pressure of a gas to 1.0 x 10-6 atm, what is the pressure expressed in millimeters of mercury
Answer:
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
Explanation:
According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:
[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]
[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequency of the output voltage. C) only the amplitude of the output voltage. D) only the phase of the output voltage.
Answer:
A) the frequency and amplitude of the output voltag
Explanation:
Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.
Changing the speed regulator will change the engine throttle setting to maintain the speed.
While the power, torque, current, fuel flow rate and torque angle will have decreased.
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
An inductor is hooked up to an AC voltage source. The voltage source has EMF V0 and frequency f. The current amplitude in the inductor is I0.
Part A
What is the reactance XL of the inductor?
Express your answer in terms of V0 and I0.
Part B
What is the inductance L of the inductor?
Express your answer in terms of V0, f, and I0.
Answer:
a. The reactance of the inductor is XL = V₀/I₀
b. The inductance of the inductor is L = V₀/2πfI₀
Explanation:
PART A
Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,
XL = V₀/I₀
So, the reactance of the inductor is XL = V₀/I₀
PART B
The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor
Since XL = V₀/I₀ = 2πfL
V₀/I₀ = 2πfL
L = V₀/2πfI₀
So the inductance of the inductor is L = V₀/2πfI₀
A) The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex]
B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
A) Expressing the Reactance of the inductor
Voltage across the Inductor = V₀ = I₀XL ---- ( 1 )
Where : V₀ = emf voltage , I₀ = current
from equation ( 1 )
∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]
B ) Expressing the Inductance of the Inductor
Inductance of an inductor is expressed as : XL = 2πfL
from part A
XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL
∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀
L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
Hence we can conclude that The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex] and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex] .
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At a rock concert, a dB meter registered 131 dB when placed 2.6 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.
a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
b) How far away would the sound level be 86 dB?
Answer:
Explanation:
A) 131 dB = 10*log(I / 1e-12W/m²)
where I is the intensity at 2.6 m away.
13.1 = log(I / 1e-12W/m²
1.25e13= I / 1e-12W/m²
I = 1.25 x10^1W/m²
power = intensity * area
P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄
B) 86 dB = 10*log(I / 1e-12W/m²)
8.6 = log(I / 1e-12W/m²)
3.98e8 = I / 1e-12W/m²
I = 3.98e-4 W/m²
area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²
A = 4πr²
2.66e6 m² = 4πr²
r = 14.5m ◄
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.
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Compare the value for the inductor when the current was increasing vs decreasing. Which statement matches the expected results. The inductance should be the same regardless of whether the current is increasing or decreasing. The inductance should be greater while the current is increasing. The inductance should be greater while the current is decreasing.
Answer:
see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,
Explanation:
The express for inductance is
[tex]E_{L}[/tex]= L dI / dt
L = E_{L} (di / dt)⁻¹
where L is the inductance, E_{L} the induced electromotive force, di/dt the variation of the current as a function of time.
When analyzing this equation we see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,
Correct answer the inductance must be the same regardless of whether the current increases or decreases.
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
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth
Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
g' = 1.09 m/s²
Answer:
The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]Explanation:
Surface of earth,
[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]
free fall acceleration at height h,
[tex]g_h = \frac{GM}{(R+h)^2}[/tex]
where
G = gravitational constant
R = Radius of earth
M = mass of earth
therefore,
[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]
Where height h = 2R
[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]
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How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
The intensity of the waves from a point source at a distance d from the source is I. What is the intensity at a distance 2d from the source?
Answer:
The intensity at distance 2d from source is [tex]I_1 = \frac{1}{4} * I[/tex]
Explanation:
From the question we are told that
The distance of the wave from point source is d
The intensity is [tex]I[/tex]
The distance we are considering is 2d
Generally the intensity of a wave is mathematically represented as
[tex]I = \frac{ P }{\pi d^2 }[/tex]
Here P is power of point source
Now when d = 2d
[tex]I_1 = \frac{ P }{\pi (2d)^2 }[/tex]
[tex]I_1 = \frac{ 1 }{4 } * \frac{ P }{\pi d^2 }[/tex]
=> [tex]I_1 = \frac{1}{4} * I[/tex]
The intensity at a distance 2d from the source is equal to [tex]I'=\frac{I}{4}[/tex]
Given the following data:
Distance = dIntensity = ITo determine the intensity at a distance 2d from the source:
Mathematically, the intensity of a wave is given by the formula:
[tex]I=\frac{P}{\pi d^2}[/tex]
Where:
I is the intensity of a wave.P is the power.d is the distance.Since the distance is doubled (2d), we have:
Let the new intensity be [tex]I'[/tex][tex]I'=\frac{P}{\pi (2d)^2}\\\\I'=\frac{P}{4\pi (d)^2}\\\\I'=\frac{1}{4} \times \frac{P}{\pi (d)^2}\\\\I'=\frac{1}{4} \times I\\\\I'=\frac{I}{4}[/tex]
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Explain how surface waves can have characteristics of both longitudinal waves and transverse waves. Please use 3 content related sentences
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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.
Explanation:
Surface waves can exhibit characteristics of both longitudinal waves and transverse waves.
Surface waves are a type of mechanical wave that propagate along the interface between two different mediums, such as the ground and air or the surface of water. These waves combine properties of both longitudinal and transverse waves
Similar to longitudinal waves, surface waves involve particles oscillating in the same direction as the wave propagation. This creates compressions and rarefactions, leading to variations in density or pressure. These compressions and rarefactions are characteristic of longitudinal waves.
However, surface waves also exhibit transverse motion. As the wave propagates along the surface, particles move in a perpendicular direction to the wave's motion. This transverse motion causes particles to displace vertically or horizontally, similar to transverse waves.
By combining both longitudinal and transverse characteristics, surface waves possess a complex motion that allows them to travel along the surface while simultaneously causing particles to oscillate both parallel and perpendicular to the direction of wave propagation.
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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?
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%
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The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.
79.70% Mg-24 (23.9872 amu), 10.13% Mg-25 (24.9886 amu), and 10.17% Mg-26 (25.9846 amu).
A. How many neutrons are in Mg-25?
B. What is the average atomic mass of magnesium in these rocks?
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
Answer:
A. number of neutrons of Magnesium Mg = 13
B. The average mass of Mg = 22.29 amu
C. the magnesium composition on Mars is not the same as that on Earth.
Explanation:
Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.
The atomic number of Magnesium Mg = 12
The atomic number of an element is the number of protons present in the atomic nucleus of the element
i.e Atomic number = number of protons = 12
The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.
Mass number = number of protons + number of neutrons
Given that the mass number of Mg = 25
Then;
25 = 12 + number of neutrons
25 - 12 = number of neutrons
13 = number of neutrons
number of neutrons of Magnesium Mg = 13
B. What is the average atomic mass of magnesium in these rocks?
The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.
Therefore the average atomic mass of magnesium can be calculated as:
= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]
= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]
= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]
The average mass of Mg = 22.29 amu
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.
There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.43 A. Required:a. Detrmine the resistance of one of the bulbs.b. What would the current be if the bulbs were connected in series?
Answer:
a)3000ohm
b)4.44mA
Explanation:
a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
But according to ohm'law which can be expressed below
V=IR
R=V/I
R(total)= 120/0.36
= 333.33ohm
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
1/R(total)=9/R
1/333.33= 9/R
R= 3000ohm
Therefore, the resistance is 3000ohm
b)the bulbs were connected in series here, then for series connection we use below expression
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
R(total)=9R
= 9*3000
=27000ohm
I=VR
I=V/R
I= 120/27000
= 4.44*10⁻³A
4.44mA
Therefore, the current is 4.4mA
When a ray of light passes from glass to water it is?
Answer:
[tex]\huge\boxed{Refracted}[/tex]
Explanation:
When a ray of light passes from glass to water, it
1) is Slightly refracted (bending of light)
2) moves away from the normal.
Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.
Answer:
refraction
Explanation:
The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature in degrees Celsius when 1.50 kcal of heat enters 1.50 kg of the following, originally at 15.0°C.(a) water
(b) concrete
(c) steel
(d) mercury
Answer:
Final temperature Water = 20.99-degree celsius
Final temperature Concrete = 24.98 degree celsius
Final temperature Steel = 50.1 degree celsius
Final temperature Mercury = 29.26 degree celsius
Explanation:
Given the mass of each substance = 1.50 kg
Ti = 15
Q = 1.5 kcal = 6276 joule
We have to use the heat capacity of each object so find the heat capacity from the table.
Heat capacity of water = 4186 J/kg degree celsius.
Heat capacity of concrete = 840 J/kg degree celsius.
Heat capacity of steel = 452 J/kg degree celsius.
Heat capacity of mercury = 139 J/kg degree celsius.
Use the below formula to find the final temperature.
[tex]T_f = T_i + \frac{Q}{mc_w} \\[/tex]
[tex]\text{Temperature in the case of water.} \\= 20 + \frac{6276}{1.5 \times 4186 } \\= 20.99 \ degree \ celsius \\\text{Temperature in the case of concrete.} \\= 20 + \frac{6276}{1.5 \times 840 } \\= 24.98 \ degree \ celsius \\\text{Temperature in the case of steel.} \\= 20 + \frac{6276}{1.5 \times 452 } \\= 29.26 \ degree \ celsius \\\text{Temperature in the case of mercury.} \\= 20 + \frac{6276}{1.5 \times 139 } \\= 50.1 \ degree \ celsius \\[/tex]
Specular reflection occurs where the light ray in the glass strikes the reflector. If no light is to enter the water, we require that there be reflection only. Which phenomenon prevents the light from entering the water?
Answer:
The critical angle phenomenon.
Explanation:
Critical angle in optics is the smallest angle of incidence of a wave, that will give total reflection of the wave. This phenomenon occurs at the boundary of two medium, where light will normally move from one medium to another.
To prevent light from entering the water, the angle of incidence of the light incident on the water must exceed the critical angle.