shows the viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum a. What will happen to the fringe spacing if the wavelength of the light is decreased? b. What will happen to the fringe spacing if the spacing between the slits is decreased? c. What will happen to the fringe spacing if the distance to the screen is decreased? d. Suppose the wavelength of the light is 500 nm. How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit?

Answers

Answer 1

The fringe spacing in a double-slit experiment decreases as the wavelength of the light decreases, the spacing between the slits decreases, and the distance to the screen decreases. The difference in path length between the dot on the screen in the center of fringe E and the left slit is (3λd)/(2θ).

a. If the wavelength of the light is decreased, the fringe spacing will decrease. This is because fringe spacing is directly proportional to the wavelength of light.

b. If the spacing between the slits is decreased, the fringe spacing will increase. This is because fringe spacing is inversely proportional to the slit spacing.

c. If the distance to the screen is decreased, the fringe spacing will increase. This is because fringe spacing is inversely proportional to the distance between the slits and the screen.

d. Using the small angle approximation, the path difference between the dot in the center of fringe E and the left slit is approximately (d/2)sin(θ). The path difference to the right slit is the same but with the opposite sign for θ. The difference in path length is approximately d sin(θ) which equals 3λ/2. Assuming sin(θ) ≈ θ, the distance to the left slit is (3λd)/(2θ).

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Related Questions

Sam is stationary and then starts skateboarding. His velocity increases to 5m/s west over a period of 10 seconds. What is Sam’s average acceleration?

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Sam's average acceleration is 0.5 m/s² west, calculated by dividing the change in velocity (5 m/s) by the time (10 s).

Sam is initially stationary and then starts skateboarding with his velocity increasing to 5 meters per second (m/s) west over a period of 10 seconds.

To find his average acceleration, we need to divide the change in velocity by the time it took for the change to occur.

In this case, Sam's change in velocity is 5 m/s (from 0 m/s to 5 m/s) and the time taken is 10 seconds.

By dividing the change in velocity (5 m/s) by the time (10 s), we find that Sam's average acceleration is 0.5 meters per second squared (m/s²) west.

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A round bottom flask contains 3.15 g of each methane, ethane, and butane is conta in ed in a 2.00 L flask at a temperature of 64 °C. a.) What is the partial pressure of each of the gases within the flask? b.) Calculate the total pressure of the mixture.

Answers

a) The partial pressure of methane is 2.49 atm, ethane is 1.33 atm, and butane is 0.68 atm.

b) The total pressure of the mixture is 4.50 atm.

To calculate the partial pressure of each gas, we can use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to find the number of moles of each gas. We can use the formula:

moles = mass / molar mass

For methane (CH4):

moles(CH4) = 3.15 g / 16.04 g/mol = 0.196 mol

For ethane (C2H6):

moles(C2H6) = 3.15 g / 30.07 g/mol = 0.105 mol

For butane (C4H10):

moles(C4H10) = 3.15 g / 58.12 g/mol = 0.054 mol

Next, we can calculate the partial pressure of each gas using the ideal gas law:

P(CH4) = (moles(CH4) * R * T) / V

P(C2H6) = (moles(C2H6) * R * T) / V

P(C4H10) = (moles(C4H10) * R * T) / V

Assuming R = 0.0821 L*atm/mol*K and converting the temperature to Kelvin (64 °C = 337 K), and the volume is given as 2.00 L, we can substitute the values to calculate the partial pressures.

For methane (CH4):

P(CH4) = (0.196 mol * 0.0821 L*atm/mol*K * 337 K) / 2.00 L = 2.49 atm

For ethane (C2H6):

P(C2H6) = (0.105 mol * 0.0821 L*atm/mol*K * 337 K) / 2.00 L = 1.33 atm

For butane (C4H10):

P(C4H10) = (0.054 mol * 0.0821 L*atm/mol*K * 337 K) / 2.00 L = 0.68 atm

To calculate the total pressure of the mixture, we sum up the partial pressures of each gas:

Total pressure = P(CH4) + P(C2H6) + P(C4H10)

Total pressure = 2.49 atm + 1.33 atm + 0.68 atm = 4.50 atm

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the surface finish produced by electrical discharge machining has a cross-hatch pattern with a roughness value of 32 to 120 microinches. T/F?

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True. The surface finish produced by electrical discharge machining (EDM) often exhibits a cross-hatch pattern and has a roughness value typically ranging from 32 to 120 microinches.

The roughness value is a measure of the irregularities or variations in the surface texture, with a higher value indicating a rougher surface. The cross-hatch pattern is a result of the electrode's movement during the EDM process, leaving characteristic lines on the surface. This pattern can help improve lubrication and wear resistance in certain applications. However, it's important to note that the specific roughness values can vary depending on various factors such as the EDM parameters, electrode material, and the workpiece material.

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A girl strikes a 0.445kg soccer ball with a net force of 5.92N. What is the acceleration of the soccer ball? 0 13.3 m/s2 O 0.0752 m/s2 0 6.36 m/s2 0 5.48 m/s2

Answers

The answer to the question is 13.3 m/s2, if a girl strikes a 0.445kg soccer ball with a net force of 5.92N.


To find the acceleration of the soccer ball, we can use the formula F = ma, where F is the net force applied to the ball, m is the mass of the ball, and a is the acceleration of the ball. We know that the mass of the ball is 0.445kg and the net force applied is 5.92N. Substituting these values into the formula, we get:

5.92N = 0.445kg x a

Solving for a, we get:

a = 5.92N / 0.445kg

a ≈ 13.3 m/s2

Therefore, the answer is that the acceleration of the soccer ball is 13.3 m/s2.

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Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is. show that the polar equation of the orbit of a planet is where is the eccentricity.

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The polar equation of the orbit of a planet with a focus at the pole, major axis on the polar axis, and length of major axis a is r = a(1-e^2)/(1+e*cos(theta)).

When a planet orbits a star, the shape of the orbit can be described using a polar equation. In this case, the focus is at the pole, which means that the planet's distance from the star is the same as the distance from the pole. The major axis lies on the polar axis, which means that the distance between the star and the farthest point on the orbit is a. Finally, the length of the major axis is related to the eccentricity of the orbit, which is the ratio of the distance between the foci to the length of the major axis.

Using these parameters, we can derive the polar equation of the orbit as r = a(1-e^2)/(1+e*cos(theta)), where r is the distance from the star to the planet, theta is the angle from the polar axis, a is the length of the major axis, and e is the eccentricity.

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the earth is approximately spherical, with a diameter of 1.27×107m1.27×107m. it takes 24.0 hours for the earth to complete one revolution.

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Answer:This statement seems incomplete. Please provide the rest of the question.

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In what direction is matter displaced in a traverse wave

A. In the same direction of the energy flow

B. In the opposite direction of the energy flow

C. In a spiral around the energy flow

D. At a right angle to the energy flow

Answers

Answer:

In a transverse wave, the matter is displaced perpendicular (at a right angle) to the direction of energy flow. Therefore, the correct answer is D.

Explanation:

a sinusoidal electromagnetic wave has rms electric field 800 n/c. what is the intensity of the wave? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 t ∙ m/a, ε0 = 8.85 × 10-12 c2/n ∙ m2)

Answers

The intensity of the wave is 1.92 x [tex]10^{-5[/tex] W/[tex]m^2[/tex].

The intensity (I) of an electromagnetic wave is defined as the average power per unit area that it carries.

The relationship between the intensity and the RMS electric field (E) of a sinusoidal electromagnetic wave is given by:

I = (1/2) x [tex]\epsilon_0[/tex] x c x [tex]E^2[/tex]

where [tex]\epsilon_0[/tex] is the vacuum permittivity and c is the speed of light in vacuum.

Substituting the given values, we have:

I = [tex](1/2) \times (8.85 \times 10^{-12}) \times (3.00 \times 10^8) \times (800 \times 10^{-9})^2[/tex]

I = 9.44 × [tex]10^{-5} W/m^2[/tex]

Therefore, the intensity of the wave is 9.44 × [tex]10^{-5[/tex] W/[tex]m^2[/tex]

It is important to note that the intensity of an electromagnetic wave depends on the square of the amplitude of its electric field.

Therefore, doubling the RMS electric field of the wave will result in a four-fold increase in its intensity.

Conversely, reducing the electric field amplitude by a factor of 2 will result in a reduction of the wave's intensity by a factor of 4.

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The intensity (I) of an electromagnetic wave is defined as the average power (P) per unit area (A) of the wave, and can be calculated using the following formula:

I = P/A

We can also express the power of an electromagnetic wave in terms of the electric field (E) and magnetic field (B) amplitudes, using the following relationship:

P = (1/2)ε0cE^2

where ε0 is the permittivity of free space, c is the speed of light, and E is the root-mean-square (rms) electric field amplitude.

Since we are given the rms electric field amplitude, we can use the above equation to calculate the power of the wave:

P = (1/2)ε0cE^2 = (1/2)(8.85 × 10^-12 c^2/n ∙ m^2)(3.00 × 10^8 m/s)(800 × 10^-9 V/m)^2 = 0.095 W/m^2

Next, we can calculate the intensity by dividing the power by the area through which the wave is passing. Since the area of a sphere of radius r is 4πr^2, and the wave is assumed to be spreading out uniformly in all directions, we can take the area to be that of a sphere with radius r = 1 meter:

A = 4πr^2 = 4π(1 m)^2 = 12.57 m^2

Therefore, the intensity of the wave is:

I = P/A = 0.095 W/m^2 ÷ 12.57 m^2 = 7.57 × 10^-3 W/m^2

So the intensity of the wave is 7.57 × 10^-3 W/m^2.

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A mass oscillates on a spring with a period of 0.89s and an amplitude of 5.9cm. Find an equation giving x as a function of time, assuming the mass starts at x=A at time t=0 .

Answers

The equation describing the motion of a mass oscillating on a spring with a period of 0.89s and an amplitude of 5.9cm, starting at x=A at time t=0, is x = 5.9cos((2π/0.89)t).

The motion of a mass on a spring can be described by the equation x = Acos(ωt + φ), where A is the amplitude of the motion, ω is the angular frequency, t is time, and φ is the phase constant. The period (T) of the motion is given by T = 2π/ω. In this case, the period is given as 0.89s, so we can calculate the angular frequency as ω = 2π/T = 7.03 rad/s.

The mass starts at x=A, so the phase constant can be found using the initial condition x(0) = A, which gives φ = 0. Substituting the values of A, ω, and φ into the equation for motion, we get x = 5.9cos(7.03t).

Therefore, the equation describing the motion of the mass is x = 5.9cos((2π/0.89)t), which gives the position of the mass as a function of time.

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10-4. calculate the required diameter for certified-capacity liquid rupture discs for the following conditions. assume a liquid specific gravity of 1.2 for all cases. Liquid flow Set pressure Overpressure Backpressure a. 500 gpm b. 100 gpm c. 5 m/s d. 10 m/s 100 psig 50 psig 10 barg 20 barg 10 psig 5 psig 1 barg 2 barg 5 psig 2 psig 0.5 barg 1 barg

Answers

The required diameter for certified-capacity liquid rupture discs for the given conditions are 6.08 inches for 500 gpm, 3.07 inches for 100 gpm, 1.29 inches for 5 m/s, and 1.60 inches for 10 m/s.

To calculate the required diameter for certified-capacity liquid rupture discs for the given conditions, we first need to determine the burst pressure for each case. The burst pressure is calculated using the following formula:
Burst Pressure = Set Pressure + Overpressure - Backpressure
Using the specific gravity of 1.2 for all cases, we can calculate the burst pressure for each scenario as follows:
a. 500 gpm: Burst Pressure = 100 psig + 50 psig - 10 psig = 140 psig
b. 100 gpm: Burst Pressure = 100 psig + 50 psig - 5 psig = 145 psig
c. 5 m/s: Burst Pressure = 10 barg + 1 barg - 0.5 barg = 10.5 barg
d. 10 m/s: Burst Pressure = 20 barg + 2 barg - 1 barg = 21 barg
Once we have the burst pressure, we can use the specific gravity and the following formula to calculate the required diameter of the rupture disc:
Diameter = (Flow Rate * 60 * Specific Gravity) / (Burst Pressure * 0.8 * 3.14)
Where:
Flow Rate = Liquid flow in gallons per minute (gpm) or meters per second (m/s)
Specific Gravity = 1.2
Burst Pressure = Calculated burst pressure in psig or barg
Using the above formula, we can calculate the required diameter for each scenario as follows:
a. 500 gpm: Diameter = (500 * 60 * 1.2) / (140 * 0.8 * 3.14) = 6.08 inches
b. 100 gpm: Diameter = (100 * 60 * 1.2) / (145 * 0.8 * 3.14) = 3.07 inches
c. 5 m/s: Diameter = (5 * 60 * 1.2) / (10.5 * 0.8 * 3.14) = 1.29 inches
d. 10 m/s: Diameter = (10 * 60 * 1.2) / (21 * 0.8 * 3.14) = 1.60 inches
Therefore, the required diameter for certified-capacity liquid rupture discs for the given conditions are 6.08 inches for 500 gpm, 3.07 inches for 100 gpm, 1.29 inches for 5 m/s, and 1.60 inches for 10 m/s.

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If a sheet containing a single thin slit is heated (without damaging it) and therefore expands, what happens to the width of the central bright diffraction region on a distant screen? A.it gets narrower B.It gets wider C. It doesnt change

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The width of the central bright diffraction region on a distant screen will get wider if a sheet containing a single thin slit is heated and expands. This is because the width of the central bright diffraction region is directly proportional to the width of the slit.  Correct answer is option B

As the slit expands due to heating, its width also increases, leading to a wider central bright diffraction region on the distant screen. This phenomenon can be explained by the principles of diffraction, which states that when a wave passes through an aperture, it diffracts and spreads out. In the case of a single thin slit, the light passing through the slit diffracts and creates a pattern of alternating bright and dark fringes on the distant screen.

The central bright fringe corresponds to the direct transmission of light through the center of the slit, and its width is dependent on the width of the slit.



Therefore, if the slit expands due to heating, the width of the central bright fringe also increases, resulting in a wider diffraction pattern on the screen. However, it is important to note that the intensity of the bright fringe decreases as the width increases, leading to a dimmer diffraction pattern overall.  Correct answer is option B

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two loudspeakers in a 20°c room emit 686hz sound waves along the x- axis. an observer is located at x0.a. if the speakers are in phase, what is the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive?b. if the speakers are out of phase, what is the smallest distance between the speakers for which the interference of the sound waves is maximum constructive?

Answers

Sure! Sound waves are vibrations that propagate through a medium, such as air, and can be described by their frequency, which is measured in hertz (Hz). Interference occurs when two or more waves overlap in space and time. If the waves are in phase, meaning their peaks and troughs align, they will create constructive interference, where the amplitude of the resulting wave is increased. If they are out of phase, meaning their peaks and troughs are misaligned, they will create destructive interference, where the amplitude of the resulting wave is decreased.

a. For destructive interference, we want the waves from the two speakers to cancel each other out. This occurs when the path difference between the waves is equal to a half-wavelength, or λ/2. The formula for wavelength is λ = v/f, where v is the speed of sound (343 m/s at 20°C) and f is the frequency (686 Hz). Therefore, λ = 343/686 = 0.5 m. The path difference between the waves at point x0 will depend on the distance between the speakers, which we'll call d. If d is the smallest distance for which we get destructive interference, then the path difference will be λ/2. Using the geometry of the situation, we can see that this occurs when sinθ = λ/(2d), where θ is the angle between the line connecting the speakers and the observer and the x-axis. Since θ = 10° (half of the 20° angle between the x-axis and the line connecting the speakers), we can solve for d: d = λ/(2sinθ) = 0.086 m.

b. For constructive interference, we want the waves from the two speakers to reinforce each other. This occurs when the path difference between the waves is equal to an integer number of wavelengths, or nλ. If the speakers are out of phase, the path difference will be λ/2 + nλ, where n is an odd integer. If the speakers are in phase, the path difference will be nλ, where n is an even integer. In either case, we want the path difference to be as small as possible, which means n should be as small as possible. Since we want constructive interference, we'll choose the smallest even integer, which is n = 2. Therefore, the path difference is 2λ = 1 m. Using the same formula as before, sinθ = nλ/(2d), we can solve for d: d = nλ/(2sinθ) = 0.214 m.

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a wheel 213 cmcm in diameter takes 2.85 ss for each revolution, find its period and angular speed in rad/s

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The period of the wheel is 2.85 seconds and its angular speed is 2.20 rad/s.

To start, we need to convert the diameter of the wheel from cm to meters, as angular speed is typically measured in radians per second and period is measured in seconds.

213 cm = 2.13 m

Next, we can use the formula for period:

Period = time for one revolution

We know that it takes 2.85 seconds for each revolution, so:

Period = 2.85 s

Now, we can use the formula for angular speed:

Angular speed = 2π / Period

We just found the period to be 2.85 seconds, so:

Angular speed = 2π / 2.85 s

Angular speed = 2.20 rad/s

Therefore, the period of the wheel is 2.85 seconds and its angular speed is 2.20 rad/s.

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Write a claim that responds to the question: Why can transferring energy into or out of a substance change molecules’ freedom of movement? Be sure to include the words kinetic energy, temperature, and speed in your response.

Answers

The claim can be that, Kinetic energy refers to the perpetual motion of molecules in a material.

A molecule's temperature and speed are exactly related to how much kinetic energy it has. Molecules acquire more kinetic energy when energy is introduced into a substance by heating, which causes them to move more quickly and raise temperature. This rise in temperature and speed may cause more frequent collisions and increased movement among molecules.

While molecules lose kinetic energy when energy is transported out of a substance through cooling, which causes them to travel more slowly and drop in temperature. The molecules may travel more slowly and experience fewer collisions as a result of the drop in temperature and speed. As a result, the freedom of motion of a substance's molecules can be significantly impacted by the passage of energy into or out of it.

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four objects are situated along the y axis as follows: a 1.92-kg object is at 3.01 m, a 2.93-kg object is at 2.42 m, a 2.53-kg object is at the origin, and a 4.05-kg object is at -0.498 m. where is the center of mass of these objects?

Answers

The center of mass of the four objects is approximately 0.95 meters along the y-axis.

To find the center of mass (COM) of the four objects along the y-axis, we will use the formula:

COM = (m1*y1 + m2*y2 + m3*y3 + m4*y4) / (m1 + m2 + m3 + m4)

Where m1, m2, m3, and m4 are the masses of the objects, and y1, y2, y3, and y4 are their respective positions on the y-axis. Plugging in the given values:

COM = ((1.92 kg * 3.01 m) + (2.93 kg * 2.42 m) + (2.53 kg * 0 m) + (4.05 kg * -0.498 m)) / (1.92 kg + 2.93 kg + 2.53 kg + 4.05 kg)

COM = ((5.7792 kg*m) + (7.0936 kg*m) + (0 kg*m) + (-2.0169 kg*m)) / (11.43 kg)

COM = (10.8559 kg*m) / (11.43 kg)

COM ≈ 0.95 m

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ASAP HELP PLEASE

The speed of light is 300,000,000 m/s, the speed of sound is 343 m/s.


If an airplane is 10 km (10,000 m), how much time difference would there be between you seeing the plane and hearing it?


DO NOT ABUSE THE POINT SYSTEM

YOU WILL BE REPORTED

Answers

The time difference between seeing the airplane and hearing it would be:

Time difference = time for light to travel - time for sound to travel

= 0.0000333 s - 29.15 s = -29.15 seconds (approx.)

This negative time difference means that we would hear the sound of the airplane before we see it, since the sound takes longer to reach us than the light.

To calculate the time difference between seeing an airplane and hearing it, we need to determine how long it takes for the sound to travel from the airplane to our ears. We can then subtract this time from the time it takes for the light to travel from the airplane to our eyes.

The distance between us and the airplane is 10,000 meters. Since sound travels at a speed of 343 m/s, we can divide the distance by the speed of sound to get the time it takes for the sound to reach us:

Time for sound to travel = distance / speed of sound = 10,000 m / 343 m/s = 29.15 seconds (approx.)

On the other hand, since light travels at a speed of 300,000,000 m/s, we can divide the distance by the speed of light to get the time it takes for the light to reach us:

Time for light to travel = distance / speed of light = 10,000 m / 300,000,000 m/s = 0.0000333 seconds (approx.)

Therefore, the time difference between seeing the airplane and hearing it would be:

Time difference = time for light to travel - time for sound to travel

= 0.0000333 s - 29.15 s = -29.15 seconds (approx.)

This negative time difference means that we would hear the sound of the airplane before we see it, since the sound takes longer to reach us than the light.

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Tall Pacific Coast redwood trees (Sequoia sempervirens) can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia cone moving when it reaches the ground if it dropped from the top of a 100 m tree?

Answers

To determine the speed at which a sequoia cone would hit the ground when dropped from the top of a 100 m tall tree, we can use the principles of free fall motion.

When air drag is negligible, the only force acting on the cone is gravity. The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s² on Earth.

The speed (v) of an object in free fall can be calculated using the equation:

v = √(2gh),

where h is the height from which the object falls. In this case, h is 100 m.

Plugging in the values:

v = √(2 * 9.8 m/s² * 100 m) ≈ √(1960) ≈ 44.27 m/s.

Therefore, the sequoia cone would be moving at approximately 44.27 meters per second (m/s) when it reaches the ground.

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An EM wave has frequency 8.59×10 14
Hz. Part A What is its wavelength? * Incorrect; Try Again; 2 attempts remaining Part B How would we classity it? infrared visible light

Answers

Part A: The wavelength of an EM wave with a frequency of 8.59×10^14 Hz is approximately 3.49×10^-7 meters.

Part B: This EM wave would be classified as visible light.

To determine the wavelength of an electromagnetic (EM) wave, you can use the formula: wavelength = speed of light / frequency. The speed of light is approximately 3.00×10^8 meters per second. Using the given frequency of 8.59×10^14 Hz, the wavelength can be calculated as follows:

Wavelength = (3.00×10^8 m/s) / (8.59×10^14 Hz) ≈ 3.49×10^-7 meters

As for the classification, the electromagnetic spectrum is divided into different regions based on wavelength or frequency. Visible light has wavelengths ranging from approximately 4.00×10^-7 meters (400 nm) to 7.00×10^-7 meters (700 nm). Since the calculated wavelength of this EM wave (3.49×10^-7 meters) falls within this range, it would be classified as visible light.

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If the highest-frequency sound you can hear is 1.3×10^4hz , then what is its period?

Answers

The period of a sound wave is the time it takes for one complete cycle of wave to occur. It is measured in seconds. To find the period of a sound wave, we can use formula: Period = 1/frequency.  Period of highest-frequency sound that can be heard is [tex]7.69×10^{-5}[/tex] seconds



Where frequency is the number of cycles per second, measured in Hertz (Hz). In this case, the highest-frequency sound that can be heard is 1.3×[tex]10^{-4}[/tex]. Therefore, the period can be calculated as:  Period = 1/1.3×[tex]10^{4}[/tex] Hz , Period = 7.69×[tex]10^{-5}[/tex] seconds This means that it takes 7.69×[tex]10^{-5}[/tex] seconds for one complete cycle of the highest-frequency sound that can be heard.



It is important to note that the human ear can only hear sounds within a certain range of frequencies, typically between 20 Hz and 20,000 Hz. Sounds with frequencies below 20 Hz are called infrasound, while sounds with frequencies above 20,000 Hz are called ultrasound. The period of these sounds will vary depending on their frequency.

In conclusion, the period of the highest-frequency sound that can be heard is 7.69×[tex]10^{-5}[/tex] seconds, which is the time it takes for one complete cycle of the sound wave to occur.

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FILL IN THE BLANK ____ satellites travel at a speed and direction that keeps pace with the earth's rotation, so they appear (from earth) to remain stationary over a given spot.

Answers

Answer:

Geosynchronous

Explanation:

Quizlet

Geostationary satellites travel at a speed and direction that keeps pace with the earth's rotation, so they appear (from earth) to remain stationary over a given spot.

A photon with a wavelength of 3.40×10−13 m strikes a deuteron, splitting it into a proton and a neutron.
A) Calculate the kinetic energy released in this interaction. (MeV)
B)Assuming the two particles share the energy equally, and taking their masses to be 1.00 u, calculate their speeds after the photodisintegration. (m/s)

Answers

The kinetic energy released in this interaction is approximately [tex]5.83 * 10^{-13} J[/tex], or 0.364 MeV, and the speeds of the proton and neutron after the photodisintegration are approximately [tex]2.84 * 10^6[/tex] m/s.

A) To calculate the kinetic energy released in this interaction, we need to find the initial and final energies of the photon and the deuteron, respectively, and then subtract them.

The initial energy of the photon can be found using the formula E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

E = hc/λ = [tex](6.626 * 10^{-34} Js * 2.998 * 10^8 m/s)/(3.40 * 10^{-13} m) = 5.83 * 10^{-13} J[/tex]

The final energies of the proton and neutron can be found using the formula E =[tex](1/2)mv^2[/tex], where m is the mass of each particle and v is their velocity.

Since the two particles share the energy equally, each will have an energy of [tex]5.83 * 10^{-13} J/2 = 2.92 * 10^{-13} J.[/tex]

The mass of a proton and a neutron is approximately 1.0073 u. Converting to kilograms, we get:

m = [tex]1.0073 u * 1.661 * 10^{-27} kg/u = 1.674 * 10^{-27} kg[/tex]

The kinetic energy of each particle is:

E = [tex](1/2)mv^2 = 2.92 * 10^{-13} J[/tex]

Solving for v, we get:

v = [tex]\sqrt{2E/m} = 2.84 * 10^6 m/s[/tex]

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The wavelength of the photon is important because it determines its energy. The shorter the wavelength, the higher the energy. In this case, the photon has a relatively short wavelength, meaning it has a high amount of energy.

The concept of energy is crucial to understanding what happens after the photodisintegration. When the deuteron splits, the two resulting particles will share the energy equally. Since their masses are equal, this means they will also have equal speeds. By calculating the energy of the photon, we can determine the amount of energy that each particle receives and from there, we can calculate their speeds.

Finally, the term "photon" refers to a packet of energy that behaves like a particle. Photons are the building blocks of light and electromagnetic radiation. They have no mass, but they do have energy, which is directly proportional to their wavelength.
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Two people of the same mass climb the same flight of stairs the first two person climbs the stairs in 25 seconds the second person does so in 35 seconds who does the greater work:
a. the 1st person since he accomplished the task first
b. the 2nd person since he did the work longer
c. both did the same amount of work
d. cannot be determeined​

Answers

(c) Both did the same amount of work. The work done is determined by the force exerted on the stairs multiplied by the distance traveled.

(c) Both did the same amount of work. The work done is determined by the force exerted on the stairs multiplied by the distance traveled. In this scenario, the force exerted by each person is their weight, which is directly proportional to their mass. Since both people have the same mass, their weight and force exerted on the stairs are equal. Additionally, since they climbed the same flight of stairs, the distance traveled is also the same for both individuals. Therefore, the work done by each person is equal. The time taken to complete the task does not affect the amount of work done, as work is independent of the duration of the activity.

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A particle moves along the x-axis so that its velocity at time t is given by v(t) = ((t^6)-(13t^4)+(12)) / (10t^3)+3 At time t = 0, the initial position of the particle is x = 7. Find the acceleration of the particle at time t = 5.1.

Answers

A particle's velocity is given by v(t) = ((t⁶)-(13t⁴)+(12)) / (10t³)+3, and its initial position is x=7 at time t=0. The acceleration of the particle at time t=5.1 is approximately -7.8 m/s².

The first step is to find the acceleration of the particle, which can be obtained by taking the derivative of the velocity function v(t) with respect to time t. Thus, we have:

a(t) = v'(t) = ((6t⁵)-(52t³)) / ((10t³)+3)²

To find the acceleration of the particle at time t = 5.1, we substitute t = 5.1 into the acceleration function to get:

a(5.1) = ((6(5.1)⁵)-(52(5.1)³)) / ((10(5.1)³)+3)²

Simplifying this expression, we get:

a(5.1) ≈ -7.8 m/s²

Therefore, the acceleration of the particle at time t = 5.1 is approximately -7.8 m/s².

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How powerful is an engine that can do 400 J of work in 10 seconds?
(Provide your answer in both "Watts" and "horsepower".)

Answers

Answer:

[tex]P=40 \ W[/tex]

Conceptual:

What is work?Work is simply the transfer of work over a displacement. Work is a Newton-meter which is called a Joule, J. Work can be calculated using the following formulas.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Equations for Work:}}\\W=F \Delta rcos(\theta) \ \text{(Constant} \ \vec F) \\W= \int\limits^{r_2}_{r_1} {Fcos(\theta)} \, dr \ \text{(Varible} \ \vec F) \end{array}\right }[/tex]

The angle "θ" is the angle between the force applied and the direction of displacement.

What is power?Power is the amount work done per second, which is a J/s, and this is clumped together to create a Watt, W. Power can be calculated using the following formula.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Power:}}\\\\P=\frac{W}{t} \end{array}\right}[/tex]

Explanation:

Given that an engine does 400 J of work in 10 seconds. Find the power of the engine.

[tex]W=400 \ J\\t=10 \ s[/tex]

Plug these values into the formula for power.

[tex]P=\frac{W}{t} \\\\\Longrightarrow P=\frac{400}{10}\\\\\therefore \boxed{P=40 \ W}[/tex]

Thus, the engines power is calculated.

calculate the burnout velocity required to transfer the probe between the vicinity of the earth and the moon's orbit using a hohmann transfer

Answers

The burnout velocity required to transfer the probe between the vicinity of the earth and the moon's orbit using a Hohmann transfer is estimated to be 3.06 km/s.

To travel between two celestial bodies, such as Earth and the Moon, a spacecraft must follow a specific trajectory that requires a certain amount of energy. The Hohmann transfer is a commonly used method for transferring a spacecraft from one circular orbit to another by using a minimum amount of energy.

To calculate the burnout velocity required to transfer the probe between the vicinity of the Earth and the Moon's orbit using a Hohmann transfer, we can use the following formula:

V = √(μ(2/r₁ - 1/a) - μ(2/r₂ - 1/a))

Where V is the burnout velocity, μ is the gravitational parameter (3.986 × 10⁵ km³/s² for Earth), r₁ is the initial radius (Earth's radius + altitude), r₂ is the final radius (Moon's radius + altitude), and a is the semi-major axis of the transfer ellipse.

Assuming the altitude of the initial orbit is 200 km above Earth's surface and the altitude of the final orbit is 100 km above the Moon's surface, we can calculate the required burnout velocity using the above formula. The semi-major axis of the transfer ellipse can be found using the following formula:

a = (r₁ + r₂) / 2

Substituting the values and solving the equations, we get the burnout velocity to be approximately 3.06 km/s.

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the first bright fringe of an interference pattern occurs at an angle of 12.5° from the central fringe when a double slit is illuminated by a 449-nm blue laser. what is the spacing of the slits?
_____m

Answers

we can use the equation for the location of the bright fringes in an interference pattern:

y = mλD/d where y is the distance from the central fringe to the mth bright fringe, λ is the wavelength of the light, D is the distance from the slits to the screen, and d is the spacing of the slits. We are given that the first bright fringe occurs at an angle of 12.5° from the central fringe, so we can use trigonometry to find y:

tan(12.5°) = y/D y = D tan(12.5°) We also know the wavelength of the light is 449 nm, or 4.49 x 10^-7 m. Plugging in these values and solving for d: y = mλD/d D tan(12.5°) = λd d = λD / tan(12.5°) d = (4.49 x 10^-7 m)(D) / tan(12.5°) We don't know the distance from the slits to the screen, D, but we can assume it's on the order of a few meters. Let's say D = 2 m: d = (4.49 x 10^-7 m)(2 m) / tan(12.5°) d ≈ 1.11 x 10^-6 m So the spacing of the slits is approximately 1.11 μm.

About Equation

Equation in science is a mathematical statement that expresses the relationship between two or more quantities. Equations can be used to describe natural phenomena, determine variable values, or solve problems. Equations usually consist of symbols that represent quantities, operators that indicate mathematical operations, and an equal sign (=) that indicates equality.

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a certain transverse wave is described by y(x,t)=bcos[2π(xl−tτ)], where b = 6.90 mm, l = 30.0 cm, and τ = 3.80×10−2 s.

Answers

The given transverse wave is described by the equation y(x,t)=6.90 mm cos[2π(x(30.0 cm)-t(3.80×10−2 s))] to provide an explanation, this equation represents the displacement of the wave at a certain point (x) and time (t). The displacement is given by wavelength (30.0 cm) and τ is the period (3.80×10−2 s) of the wave.

The argument inside the cosine function represents the phase difference between the wave at two different points in space and time. As the wave propagates, this phase difference changes, causing the wave to oscillate with a certain frequency and wavelength. Overall, the equation y(x,t)=6.90 mm cos[2π(x(30.0 cm)-t(3.80×10−2 s))] describes the displacement of a transverse wave with a wavelength of 30.0 cm and a period of 3.80×10−2 s at a certain point (x) and time (t) transverse wave described by the equation y(x,t) = bcos[2π(x/l - t/τ)], where b = 6.90 mm, l = 30.0 cm, and τ = 3.80×10^-2 s.

The wave function for this transverse wave is y(x,t) = 6.90 mm * cos[2π(x/(30.0 cm) - t/(3.80×10^-2 s))]. 1. The given wave function is y(x,t) = bcos[2π(x/l - t/τ)]. 2. You have been given the values for b, l, and τ: b = 6.90 mm, l = 30.0 cm, and τ = 3.80×10^-2 s. 3. Replace the variables b, l, and τ with their respective values in the equation y(x,t) = 6.90 mm cos[2π(x/(30.0 cm) - t/(3.80×10^-2 s))].Now, you have the wave function for the given transverse wave with the provided values.

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The isoelectric point, pl, of the protein yeast alcohol dehydrogenase is 5.4 , while that of adenosine deaminase is 4.85 What is the net charge of yeast alcohol dehydrogenase at pH 4.6 ?L What is the net charge of adenosine deaminase at pH 3.5 ? The isoelectric point of leucine is 5.98 ; alanine , 6.01. During paper electrophoresis at pH 5.5, toward which electrode does leucine migrate? During paper electrophoresis at pH 4.3 , toward which electrode does alanine migrate? | The isoelectric point, pl, of the protein superoxide dismutase is 4.95, while that of glyceraldehyde-3-phosphate dehydrogenase is 6.55. What is the net charge of superoxide dismutase at pH 4.3 ? What is the net charge of glyceraldehyde-3-phosphate dehydrogenase at pH 6.1 ? The isoelectric point of asparagine is 5.41 ; threonine , 5.6. During paper electrophoresis at pH 6.5, toward which electrode does asparagine migrate? During paper electrophoresis at pH 4.5, toward which electrode does threonine migrate?

Answers

For adenosine deaminase: the protein has a net charge of approximately +2 at pH 3.5.

For leucine: It will not migrate towards either electrode during electrophoresis.

For alanine: It will migrate towards the negative electrode during electrophoresis.

For superoxide dismutase: the protein has a net charge of approximately +1 at pH 4.3.

For glyceraldehyde-3-phosphate dehydrogenase: the protein has a net charge of approximately -0.25 at pH 6.1.

To calculate the net charge of a protein at a certain pH, we need to compare the pH to the protein's isoelectric point (pI). At the pI, the protein has a net charge of zero. At pH values below the pI, the protein is positively charged, and at pH values above the pI, the protein is negatively charged.

For yeast alcohol dehydrogenase:

At pH 4.6, which is below the pI of 5.4, the protein is positively charged. The net charge can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pKa is the dissociation constant of the protein, [A-] is the concentration of the deprotonated form of the protein, and [HA] is the concentration of the protonated form of the protein.

Assuming a pKa of 6.0 for yeast alcohol dehydrogenase, we can write:

4.6 = 6.0 + log([A-]/[HA])

Solving for [A-]/[HA], we get:

[A-]/[HA] = 0.00316

This means that the ratio of deprotonated to protonated forms of the protein is 0.00316. Since the deprotonated form has a negative charge, we can assume that the protein has a net charge of approximately +1 at pH 4.6.

For adenosine deaminase:

At pH 3.5, which is below the pI of 4.85, the protein is positively charged. Using the same method as above, assuming a pKa of 6.0 for adenosine deaminase, we get:

3.5 = 6.0 + log([A-]/[HA])

[A-]/[HA] = 0.001

This means that the protein has a net charge of approximately +2 at pH 3.5.

For leucine:

At pH 5.5, which is between the pI values of leucine (5.98) and alanine (6.01), we need to look at the side chain of the amino acid. Leucine has a non-polar side chain and is therefore uncharged at this pH. It will not migrate towards either electrode during electrophoresis.

For alanine:

At pH 4.3, which is below the pI of alanine (6.01), the protein is positively charged. It will migrate towards the negative electrode during electrophoresis.

For superoxide dismutase:

At pH 4.3, which is below the pI of 4.95, the protein is positively charged. Using the same method as above, assuming a pKa of 6.0 for superoxide dismutase, we get:

4.3 = 6.0 + log([A-]/[HA])

[A-]/[HA] = 0.00316

This means that the protein has a net charge of approximately +1 at pH 4.3.

For glyceraldehyde-3-phosphate dehydrogenase:

At pH 6.1, which is above the pI of 6.55, the protein is negatively charged. Using the same method as above, assuming a pKa of 6.0 for glyceraldehyde-3-phosphate dehydrogenase, we get:

6.1 = 6.0 + log([A-]/[HA])

[A-]/[HA] = 1.25

This means that the protein has a net charge of approximately -0.25 at pH 6.1.

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Isoelectric point (pI) is the pH at which a molecule has no net charge and is stationary during electrophoresis. At pH values lower than pI, molecules will have a net positive charge, while at higher pH values, they will have a net negative charge. When given the pI and pH values, we can determine the net charge of a protein.

For example, yeast alcohol dehydrogenase will have a net positive charge at pH 4.6, while adenosine deaminase will have a net negative charge at pH 3.5. During paper electrophoresis, the direction in which a molecule migrates towards an electrode depends on its charge. For example, leucine will migrate towards the negative electrode at pH 5.5, while alanine will migrate towards the positive electrode at pH 4.3. Lastly, knowing the pI and pH values, we can determine the net charge of a protein, such as superoxide dismutase having a net positive charge at pH 4.3, and glyceraldehyde-3-phosphate dehydrogenase having a net negative charge at pH 6.1.

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The heat exchanger in problem 1 is a parallel-flow concentric tube heat exchanger. Hint: note the temperature changes of cold and hot fluids. True or False

Answers

True

The statement suggests that in problem 1, there are temperature changes in both the hot and cold fluids that flow through a parallel-flow concentric tube heat exchanger.

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Consider an infinite parallel-plate capacitor, with the lower plate (at z = - d / 2 ) carrying surface charge density -o, and the upper plate (at z = d / 2 ) carrying charge density +o.
(a) Determine all nine elements of the stress tensor, in the region between the plates. Display your answer as a 3 * 3 matrix:
(b) Use Eq. 8.21 to determine the electromagnetic force per unit area on the top plate. Compare Eq. 2.51.
(c) What is the electromagnetic momentum per unit area, per unit time, crossing the xy plane (or any other plane parallel to that one, between the plates)?
(d) Of course, there must be mechanical forces holding the plates apart-perhaps the capacitor is filled with insulating material under pressure. Suppose we sud- denly remove the insulator; the momentum flux (c) is now absorbed by the plates, and they begin to move. Find the momentum per unit time delivered to the top plate (which is to say, the force acting on it) and compare your answer to (b). [Note: This is not an additional force, but rather an alternative way of calculating the same force-in (b) we got it from the force law, and in (d) we do it by conservation of momentum.]

Answers

(a) The stress tensor elements for an infinite parallel-plate capacitor in the region between the plates are:

Txx = Tyy = (ε/2)E²Tzz = -TxxTxy = Tyx = Txz = Tzx = Tzy = Tyz = 0

(b) Using Eq, the electromagnetic force per unit area on the top plate is:

F = ε/2 * E² = Tzz

Comparing with Eq. 2.51, the electromagnetic force per unit area is equal to the energy density per unit volume.

(c) The electromagnetic momentum per unit area, per unit time, crossing the xy plane (or any other plane parallel to that one, between the plates) is zero, as there is no magnetic field in this region.

(d) The momentum per unit time delivered to the top plate when the insulator is removed is also equal to Tzz, which is the force acting on the top plate.

(b) The element responsible for the pressure on the top plate is Tzz, which is negative and equal in magnitude to Txx and Tyy, indicating that there is a compressive force acting in the z-direction.

(a) The stress tensor is a 3x3 matrix that describes the stress and strain in a material. For an infinite parallel-plate capacitor in the region between the plates, the stress tensor has the following elements:

Txx = Tyy = (ε/2)E², which represents the pressure acting in the x and y directions due to the electric field between the plates.

Tzz = -Txx, which represents the compressive force acting in the z-direction due to the pressure difference between the plates.

Txy = Tyx = Txz = Tzx = Tzy = Tyz = 0, which indicates that there are no shear forces acting between the plates.

(b) The electromagnetic force per unit area on the top plate is given by the negative of the diagonal element Tzz of the stress tensor, which is equal to ε/2 * E². This is in agreement with Eq. 2.51, which states that the electromagnetic force per unit area is equal to the energy density per unit volume.

(c) There is no magnetic field between the plates, so the electromagnetic momentum per unit area, per unit time, crossing any plane parallel to the plates is zero.

(d) The momentum per unit time delivered to the top plate when the insulator is removed is equal to Tzz, which is the force acting on the top plate. This is consistent with the result obtained in part (b), which shows that the electromagnetic force per unit area on the top plate is also equal to Tzz.

(b) The element responsible for the pressure on the top plate is Tzz, which is negative and equal in magnitude to Txx and Tyy. This indicates that there is a compressive force acting in the z-direction due to the pressure difference between the plates.

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