- Aldiffraction grating has 2000 lines per centimeter. At what angle will the third-order maximum (m 3) be for 520 nm wavelength green light? 1 nm = 1 x 10-nm, 1 cm=1 x 10-2 m. O 12.20 0 14.20 O 16.2 O 18.2°

Answers

Answer 1

The angle at which the third-order maximum (m = 3) will be observed for 520 nm wavelength green light is 16.2° (option C).

The expression to calculate the angular position of a given-order diffraction maximum is: Sin θ = (mλ)/a, Where, λ = wavelength of light, a = line spacing and m = order of the maximum.

So the given problem is of diffraction grating with line spacing 'a' of 2000 lines/cm for a green light with a wavelength of 520 nm. Using the above expression, the angle (θ) can be calculated as follows:

Sin θ = (mλ)/a => θ = sin⁻¹((mλ)/a)

Where, λ = 520 nm = 520 x 10⁻⁹ m and a = 1/2000 cm = 5 x 10⁻⁵ m. Third-order maximum (m = 3),

θ = sin⁻¹((3λ)/a)θ = sin⁻¹((3 × 520 x 10⁻⁹ m)/(5 x 10⁻⁵ m))

θ = 16.2°

Hence, option C is the correct answer.

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Related Questions

A light ray initially in water (n=1.33) enters a transparent substance at an angle of incidence of 42.0 ∘ , and the transmitted ray is refracted at an angle of 27.5 ∘
. Find the refractive index of the substance.

Answers

The refractive index of a transparent substance when a light ray initially in water (n=1.33) enters it at an angle of incidence of  [tex]42.0^{0}[/tex] and the transmitted ray is refracted at an angle of [tex]27.5.0^{0}[/tex] can be calculated using Snell's law.

The formula is as follows:

[tex]n_1 sin θ1 = n_2 sin θ_2[/tex]

where n1 is the refractive index of the incident medium, θ_1 is the angle of incidence, n_2 is the refractive index of the refracted medium, and θ_2 is the angle of refraction.

From the given problem,

[tex]n_1 = 1.33, θ_1 = 42.0^{∘}, and θ_2 = 27.5 ^{∘}.[/tex]

Let's substitute the given values into the equation to find n2:

[tex]n1 sin θ_1 = n_2 sin θ_2\\⇒ n_2 = (n_1 sin θ_1) / sin θ_2\\= (1.33 × sin 42.0^{∘}) / sin 27.5^{∘}≈ 2.22[/tex]

Therefore, the refractive index of the transparent substance is approximately 2.22.

In this question, you only need to give a numerical answer without any unit because the refractive index is a unitless quantity.

Hence, the answer is:n2 ≈ 2.22.

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An evacuated tube uses an accelerating voltage of
3.100E−1MegaVolts to accelerate protons to hit a copper plate.
Non-relativistically, what would be the maximum speed of these
protons?

Answers

The maximum speed of the protons accelerated by the evacuated tube is approximately 2.188 x 10⁷ m/s.

To determine the maximum speed of protons accelerated by an evacuated tube with a given voltage, we can use the equation for the kinetic energy of a non-relativistic particle:

K.E. = (1/2)mv²

where K.E. is the kinetic energy, m is the mass of the proton, and v is the velocity of the proton.

Given:

Voltage (V) = 3.100E−1 MegaVolts = 3.100E5 Volts (converted to SI units)

To find the velocity (v), we need to equate the kinetic energy to the work done by the electric field:

K.E. = eV

where e is the elementary charge (1.602E-19 Coulombs).

Now, we can solve for v:

(1/2)mv² = eV

Rearranging the equation:

v² = (2eV)/m

Taking the square root of both sides:

v = √((2eV)/m)

Substituting the known values:

v = √((2 × 1.602E-19 C × 3.100E5 V) / (1.6726219E-27 kg))

Calculating the expression:

v ≈ 2.188 x 10⁷ m/s

Therefore, the maximum speed of the protons accelerated by the evacuated tube is approximately 2.188 x 10⁷ m/s.

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A standing wave is set up on a string of length L, fixed at both ends. If 4-loops are observed when the wavelength is a = 1.5 m, then the length of the string is:

Answers

A standing wave is set up on a string of length L, fixed at both ends. If 4-loops are observed when the wavelength is a = 1.5 m, the length of the string is 3 meters.

In a standing wave on a string fixed at both ends, the number of loops or antinodes (points of maximum amplitude) is related to the wavelength and the length of the string.

The relationship between the number of loops (n), the wavelength (λ), and the length of the string (L) is given by the equation:

n = 2L/λ

In this case, you mentioned that 4 loops are observed when the wavelength is 1.5 m. We can substitute these values into the equation and solve for the length of the string (L):

4 = 2L/1.5

To find L, we can rearrange the equation:

2L = 4 × 1.5

2L = 6

L = 6/2

L = 3 meters

Therefore, the length of the string is 3 meters.

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The maximum speed a car can drive in a circle without sliding is limited by the friction force between tire and road surface. The coefficient of static friction between car tire and a circular track is 0.97. How long does it take a 2000-kg car to complete one circle if the car is driving at 85% of the maximum speed around this 100 m radius track? (Hint: find the maximum speed
first.) Is the answer different if the car mass is 3000 kg? Why?

Answers

It takes approximately 225.6 s for the 2000-kg car to complete one circle around this 100 m radius track and if the car mass is 3000 kg , then the maximum speed is different because the maximum speed a car can drive in a circle without sliding is independent of the car's mass. This is because the gravitational force on the car is balanced by the normal force from the road surface, which is proportional to the car's mass.

(a) The maximum speed for a car to drive in a circle without sliding is given as follows : Vmax=√(μRg)

where μ is the coefficient of static friction, R is the radius of the circle, and g is the acceleration due to gravity.

So, we can substitute the given values to find

Vmax =√(0.97×100×9.8) = 31.05m/s

Now we can use the following equation to find the time it takes for the 2000-kg car to complete one circle :

T = 2πr/v = 2πr/(0.85×Vmax) where r is the radius of the circle.

We can substitute the given values and solve for T :

T=2π(100)/(0.85×31.05) = 225.6 s

Thus, it takes approximately 225.6 s for the 2000-kg car to complete one circle around this 100 m radius track.

(b) The answer is different if the car mass is 3000 kg because the maximum speed a car can drive in a circle without sliding is independent of the car's mass. This is because the gravitational force on the car is balanced by the normal force from the road surface, which is proportional to the car's mass.

Therefore, the answer to the previous part of the question remains the same regardless of the car's mass.

Thus, the correct answers are (a) 225.6 s (b) if the car mass is 3000 kg , then the maximum speed is different .

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Consider a series RLC circuit having the parameters R=200Ω L=663mH , and C=26.5µF. The applied voltage has an amplitude of 50.0V and a frequency of 60.0Hz. Find (d) the maximum voltage ΔVL across the inductor and its phase relative to the current.

Answers

The maximum voltage [tex]ΔVL[/tex]across the inductor is approximately 19.76V, and its phase relative to the current is 90 degrees.

To find the maximum voltage [tex]ΔVL[/tex]across the inductor and its phase relative to the current, we can use the formulas for the impedance of an RLC circuit.

First, we need to calculate the angular frequency ([tex]ω[/tex]) using the given frequency (f):

[tex]ω = 2πf = 2π * 60 Hz = 120π rad/s[/tex]

Next, we can calculate the inductive reactance (XL) and the capacitive reactance (XC) using the formulas:

[tex]XL = ωL = 120π * 663mH = 79.04Ω[/tex]
[tex]XC = 1 / (ωC) = 1 / (120π * 26.5µF) ≈ 0.1Ω[/tex]
Now, we can calculate the total impedance (Z) using the formulas:

[tex]Z = √(R^2 + (XL - XC)^2) ≈ 200Ω[/tex]

The maximum voltage across the inductor can be calculated using Ohm's Law:

[tex]ΔVL = I * XL[/tex]

We need to find the current (I) first. Since the applied voltage has an amplitude of 50.0V, the current amplitude can be calculated using Ohm's Law:

[tex]I = V / Z ≈ 50.0V / 200Ω = 0.25A[/tex]

Substituting the values, we get:

[tex]ΔVL = 0.25A * 79.04Ω ≈ 19.76V[/tex]

The phase difference between the voltage across the inductor and the current can be found by comparing the phase angles of XL and XC. Since XL > XC, the voltage across the inductor leads the current by 90 degrees.

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Light sails gain momentum from photons. However, photons have no mass. Explain how this is possible and the principles behind this.

Answers

Light sails gain momentum from photons through the transfer of momentum, despite photons having no mass. The energy associated with photons allows them to possess momentum, which is transferred to the light sail upon collision. This transfer follows the principles of conservation of momentum, similar to billiard ball collisions. The phenomenon is explained by the principles of electromagnetic radiation and the relativistic definition of momentum.

The phenomenon of light sails gaining momentum from photons, despite photons having no mass, is explained by the principles of electromagnetic radiation and the transfer of momentum.

Photons are particles of light and are considered to be massless. However, they do possess energy and momentum. According to Einstein's theory of relativity, the energy (E) of a photon is related to its frequency (f) by the equation E = hf, where h is Planck's constant.

In classical physics, momentum (p) is defined as mass (m) multiplied by velocity (v). However, in relativistic physics, momentum can also be defined as the ratio of energy (E) to the speed of light (c). Therefore, the momentum (p) of a photon can be expressed as p = E/c.

Since photons travel at the speed of light (c), their momentum (p) is non-zero, despite having no mass. This is due to the energy associated with the photon.

When a photon collides with an object, such as a light sail, it transfers its momentum to the object. The object absorbs the momentum of the photon, resulting in a change in its velocity or direction.

The transfer of momentum from photons to the light sail follows the principles of conservation of momentum. The total momentum of the system (photon + light sail) remains conserved before and after the interaction. Therefore, the photon imparts its momentum to the light sail, causing it to gain momentum and accelerate.

This process is similar to a billiard ball collision, where the momentum of one ball is transferred to another upon collision, even though the individual balls have different masses.

In summary, light sails gain momentum from photons through the transfer of momentum, even though photons have no mass. The energy associated with photons allows them to possess momentum, and this momentum is transferred to the light sail, causing it to accelerate.

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An LRC ac series circuit with R= 20, L= 25 mH, and C= 30 pF, is attached to a 100-V (rms) ac power supply. The frequency of the power supply is adjusted so that the circuit is in resonance. Please enter number only, for example if the value is 300 watts, please enter 300, do not use scientific notation here. (a) What is the rms current in the circuit (b) What is the power dissipated by the circuit ?

Answers

(a) The rms current in the circuit is 5 Amperes.

(b) The power dissipated by the circuit is 500 Watts.

To calculate the rms current and power dissipated by the LRC series circuit, we can use the following formulas:

(a) The rms current (I) can be calculated using the formula:

I = V / Z

where V is the voltage of the power supply and Z is the impedance of the circuit.

For a series LRC circuit in resonance, the impedance (Z) can be calculated as:

Z = R

where R is the resistance in the circuit.

Substituting the given values:

I = 100 V / 20 Ω

Evaluating this expression:

I = 5 A

Therefore, the rms current in the circuit is 5 Amperes.

(b) The power dissipated by the circuit can be calculated using the formula:

P = I² × R

where P is the power dissipated and R is the resistance in the circuit.

Substituting the given values:

P = (5 A)² × 20 Ω

Evaluating this expression:

P = 500 W

Therefore, the power dissipated by the circuit is 500 Watts.

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A ball of mass 0.5 kg is moving to the right at 1 m/s, collides
with a wall and rebounds to the left with a speed of 0.8 m/s.
Determine the impulse that the wall gave the ball.

Answers

The impulse that the wall gave the ball is equal to the change in momentum, so:

Impulse = Change in momentum = -0.9 kg m/s

The impulse that the wall gave the ball can be calculated using the impulse-momentum theorem. The impulse-momentum theorem states that the impulse exerted on an object is equal to the change in momentum of the object. Mathematically, this can be written as:

Impulse = Change in momentum

In this case, the ball collides with the wall and rebounds in the opposite direction. Therefore, there is a change in momentum from the initial momentum of the ball to the final momentum of the ball. The change in momentum is given by:

Change in momentum = Final momentum - Initial momentum

The initial momentum of the ball is:

Initial momentum = mass x velocity = 0.5 kg x 1 m/s = 0.5 kg m/s

The final momentum of the ball is:

Final momentum = mass x velocity

= 0.5 kg x (-0.8 m/s) = -0.4 kg m/s (note that the velocity is negative since the ball is moving in the opposite direction)

Therefore, the change in momentum is:

Change in momentum = -0.4 kg m/s - 0.5 kg m/s = -0.9 kg m/s

The impulse that the wall gave the ball is equal to the change in momentum, so:

Impulse = Change in momentum = -0.9 kg m/s

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Achet of 100 m from the surface of the earth (Neglect the air friction) Dende of the gravitational force exerted on it by the earth the con due to privity as 9.8 m/s No need to write the unit. Please write the answer in one decimal place, (e.g.

Answers

The gravitational force experienced by the object 100 m above the surface of the Earth is 980.0 N.

To calculate the gravitational force experienced by an object, we can use the formula F = mg, where F is the force, m is the mass of the object, and g is the acceleration due to gravity. In this case, the object is 100 m above the surface of the Earth, and we need to neglect air friction. The value of g is approximately 9.8 m/[tex]s^2[/tex]. Therefore, the gravitational force is F = mg = (m)(9.8) = 980.0 N.

When an object is at a certain height above the Earth's surface, it is still within the Earth's gravitational field. The force of gravity pulls the object towards the center of the Earth. As the object moves higher, the gravitational force decreases because the distance between the object and the Earth's center increases. In this case, the object is 100 m above the surface of the Earth. By neglecting air friction, we can focus solely on the gravitational force.

Applying the formula F = mg, where m represents the mass of the object and g is the acceleration due to gravity, we can calculate the gravitational force. Since the mass of the object is not specified in the question, we cannot determine its exact value. However, we can conclude that at a height of 100 m, the gravitational force experienced by the object is 980 N, considering g to be 9.8 m/[tex]s^2[/tex].

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How high would the level be in an alcohol barometer at normal atmospheric pressure? Give solution with three significant numbers.

Answers

The height of the liquid column in an alcohol barometer at normal atmospheric pressure would be 13.0 meters

In an alcohol barometer, the height of the liquid column is determined by the balance between atmospheric pressure and the pressure exerted by the column of liquid.

The height of the liquid column can be calculated using the equation:

h = P / (ρ * g)

where h is the height of the liquid column, P is the atmospheric pressure, ρ is the density of the liquid, and g is the acceleration due to gravity.

For alcohol barometers, the liquid used is typically ethanol. The density of ethanol is approximately 0.789 g/cm³ or 789 kg/m³.

The atmospheric pressure at sea level is approximately 101,325 Pa.

Substituting the values into the equation, we have:

h = 101,325 Pa / (789 kg/m³ * 9.8 m/s²)

Calculating the expression gives us:

h ≈ 13.0 m

Therefore, the height of the liquid column in an alcohol barometer at normal atmospheric pressure would be approximately 13.0 meters.

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The best range must be used to measure a 1.2 V battery is A. 2V B. 20V C 200V D 200 mV

Answers

To measure a 1.2 V battery, the best range to use would be the 2V range. This range provides an appropriate scale for accurately measuring the voltage of the battery without overloading the instrument or losing precision.

When selecting the range for measuring a voltage, it is important to choose a range that is closest to the expected voltage value while still allowing some headroom for fluctuations and accuracy.

Using a range that is too high may result in a less precise measurement, while using a range that is too low may cause the instrument to overload and potentially damage the circuit.

In this case, since the battery voltage is 1.2 V, the 2V range is the most suitable option. It provides a range that is higher than the battery voltage, allowing for accurate measurement while maintaining precision.

Choosing a higher range, such as 20V or 200V, would result in a less precise reading due to the instrument's lower resolution and potential for increased noise.

The 200 mV range, on the other hand, is too low for measuring a 1.2 V battery, as it would likely result in an overload condition and potentially damage the measurement instrument.

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In 1-2 sentences, explain why the emission spectra of elements show lines of different colors but only in narrow bands. (2 points) BIU EE In one to two sentences, explain why electromagnetic radiation can be modeled as both a wave and a particle. (2 points) BIU 18

Answers

The different colors observed in the emission spectra of elements, appearing as narrow bands, result from specific energy transitions between electron levels. Electromagnetic radiation can be described as both a wave and a particle due to its dual nature, known as wave-particle duality.

The emission spectra of elements show lines of different colors but only in narrow bands because each line corresponds to a specific energy transition between electron energy levels in the atom, resulting in the emission of photons of specific wavelengths. Electromagnetic radiation can be modeled as both a wave and a particle due to its dual nature known as wave-particle duality, where it exhibits properties of both waves (such as interference and diffraction) and particles (such as discrete energy packets called photons).

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(1 points) Question 11 Shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which lie in the plane of the page. The wire is parallel to the long sides of the loop and is 0.50 m away from the closer side. At an instant when the emf induced in the loop is 2.0 V, what is the time rate of change of the current in the wire? Image size: S M L Max 0.50 m 0.50 m 30m Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 Enter answer here A/s

Answers

To determine the time rate of change of current in the wire, we can apply Faraday's law of electromagnetic induction. Given that the emf induced in the loop is 2.0 V, and considering the geometry of the setup, we can calculate the time rate of change of current in the wire using the formula ΔI/Δt = -ε/L, where ΔI/Δt is the time rate of change of current, ε is the induced emf, and L is the self-inductance of the wire.

According to Faraday's law of electromagnetic induction, the induced emf in a circuit is equal to the negative rate of change of magnetic flux through the circuit. In this case, the magnetic field generated by the current in the wire passes through the loop, inducing an emf in the loop.

To calculate the time rate of change of current in the wire, we can use the formula ΔI/Δt = -ε/L, where ε is the induced emf and L is the self-inductance of the wire. The self-inductance depends on the geometry of the wire and is a property of the wire itself.

Given that the induced emf in the loop is 2.0 V, and assuming the self-inductance of the wire is known, we can substitute these values into the formula to calculate the time rate of change of current in the wire in units of A/s.

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Compared to ultraviolet, gamma rays have ____ frequency, ____ wavelength, and ____ speed.
A. lower; longer; identical
B. higher; shorter; identical
C. higher; longer; higher
D. lower; shorter; lower

Answers

Compared to ultraviolet, gamma rays have higher frequency,shorter  wavelength, and identical speed. So, the correct option is option B.

what is wavelength?

Wavelength is a fundamental concept in physics and refers to the distance between successive peaks or troughs of a wave. In other words, it is the length of one complete cycle of a wave. It is usually denoted by the Greek letter lambda (λ) and is measured in units such as meters (m), nanometers (nm), or angstroms (Å), depending on the scale of the wave being considered.

In the context of electromagnetic waves, such as light, ultraviolet, and gamma rays, wavelength represents the distance between two consecutive points of the wave with the same phase, such as two adjacent crests or two adjacent troughs. Shorter wavelengths correspond to higher frequencies and higher energy, while longer wavelengths correspond to lower frequencies and lower energy.

Compared to ultraviolet waves, gamma rays have a higher frequency, shorter wavelength, and the same speed (which is the speed of light in a vacuum, denoted as "c").

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D Question 10 The self-inductance of a solenoid increases under which of the following conditions? Only the cross sectional area is decreased. Only the number of coils per unit length is decreased. Only the number of coils is increased. Only the solenoid length is increased. 1 pts

Answers

The self-inductance of a solenoid increases under the following conditions:

Increasing the number of turns

Increasing the length of the solenoid

Decreasing the cross-sectional area of the solenoid

Self-inductance is the property of an inductor that resists changes in current flowing through it. It is measured in henries.

The self-inductance of a solenoid can be increased by increasing the number of turns, increasing the length of the solenoid, or decreasing the cross-sectional area of the solenoid.

The number of turns in a solenoid determines the amount of magnetic flux produced when a current flows through it. The longer the solenoid, the more magnetic flux is produced.

The smaller the cross-sectional area of the solenoid, the more concentrated the magnetic flux is.

The greater the magnetic flux, the greater the self-inductance of the solenoid.

Here is a table that summarizes the conditions under which the self-inductance of a solenoid increases:

Condition                                  Increases self-inductance

Number of turns                                Yes

Length                                                   Yes

Cross-sectional area                                   No

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17. In experiment 10, a group of students found that the
moment of inertia of the plate+disk was 1.74x10-4 kg m2, on the
other hand they found that the moment of inertia of the plate was
0.34x10-4 kg

Answers

The main answer is that the moment of inertia of the disk in this configuration can be calculated by subtracting the moment of inertia of the plate from the total moment of inertia of the plate+disk.

To understand this, we need to consider the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to changes in its rotational motion and depends on its mass distribution. When a plate and disk are combined, their moments of inertia add up to give the total moment of inertia of the system.

By subtracting the moment of inertia of the plate (0.34x10-4 kg m2) from the total moment of inertia of the plate+disk (1.74x10-4 kg m2), we can isolate the moment of inertia contributed by the disk alone. This difference represents the disk's unique moment of inertia in this particular configuration.

The experiment demonstrates the ability to determine the contribution of individual components to the overall moment of inertia in a composite system. It highlights the importance of considering the distribution of mass when calculating rotational properties and provides valuable insights into the rotational behavior of objects.

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wheel of radius 0.35m freely rotating kicks a water droplet 52 cm into the air.
If the angular
acceleration of the wheel is -0.35 rad/s?, how many times will the wheel rotate before coming to a complete
stop?

Answers

Using the concept of Conservation of energy, the wheel will complete 0.0876 rotations before coming to a complete stop.

Given:

           Radius of the wheel, r = 0.35m

           Height of the water droplet, h = 52cm

                                                              = 0.52m

          Angular acceleration, α = -0.35 rad/s

Let n be the number of rotations required for the wheel to stop.

Concepts used: For a freely rotating wheel, the work done is zero.

Conservation of energy.

Wheel makes a full rotation when a distance equal to the circumference of the wheel has been covered.

Solution:

Work done by the wheel is zero.

∴ Change in Kinetic Energy + Change in Potential Energy = 0

In the initial state, the droplet is at the lowest point, so there is no PE.

∴ Change in KE = 0

We know,

                 KE = 0.5 Iω²

                  I is moment of inertia

                 ω is the angular velocity of the wheel.

At the maximum height, the wheel will have zero velocity, so the KE is zero.

∴ KE_initial = KE_final

   0.5 I ω_i² = 0

       Iω_i² = 0

         ω_i = 0

The work done by the wheel is zero.

∴ Change in PE + Change in KE = 0

We know,

               PE = mgh

               m is the mass of the water droplet

               h is the height at which it reaches.

       ∴ mgh = 0.5 Iω_f²

          mgh = 0.5 × (mr²) × ω_f²

              h = 0.5 r² ω_f²g

We know,

            α = ω_f / t_fα

               = -0.35 rad/s

         t_f = ω_f / α

     ∴ t_f = -ω_f / α

Substitute ω_f from above equation.

          t_f = -2h / rαg

       ∴ t_f = -2(0.52) / (0.35) × (-0.35) × (9.8)

       ∴ t_f = 1.584 s

The time taken for one complete rotation,

                T = 2π / ω_f

             ∴ T = 2π / (0.35 × 1)

             ∴ T = 18.08 s

The total number of rotations, n = t_f / T

                                              ∴ n = 1.584 / 18.08

                                                ∴ n = 0.0876 times

Thus, the wheel will complete 0.0876 rotations before coming to a complete stop.

Hence, the conclusion of this problem is that the wheel will complete 0.0876 rotations before coming to a complete stop.

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The wheel will rotate one complete revolution before coming to a complete stop.

To solve this problem, we can use the kinematic equation for angular motion:

θ = ω_initial * t + (1/2) * α * t^2

Where:

θ is the angular displacement (in radians)

ω_initial is the initial angular velocity (in rad/s)

α is the angular acceleration (in rad/s^2)

t is the time (in seconds)

Given:

Initial angular velocity, ω_initial = 0 (since the wheel starts from rest)

Angular acceleration, α = -0.35 rad/s^2

Angular displacement, θ = 2π radians (one complete rotation)

We can rearrange the equation to solve for time:

θ = (1/2) * α * t^2

t^2 = (2 * θ) / α

t = √((2 * θ) / α)

Substituting the given values, we have:

t = √((2 * 2π) / -0.35)

Calculating this, we get:

t ≈ 7.82 seconds

Now, to find the number of rotations, we can divide the angular displacement by 2π (the angle for one full rotation):

Number of rotations = θ / (2π)

Number of rotations = 2π / (2π)

Calculating this, we get:

Number of rotations = 1

Therefore, the wheel will rotate one complete revolution before coming to a complete stop.

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A diffraction grating contains 500 lines per millimetre. For normally incident light (i = 0) of wavelength 550 nm, at what angle would a first order diffraction maximum be observed. For which angles would a first order diffraction maximum be observed when light is incident at i = 15°?

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Therefore, for light incident at an angle of 15°, a first-order diffraction maximum would be observed at an angle of approximately 23.75°.

To determine the angle at which a first-order diffraction maximum is observed using a diffraction grating, we can use the formula:

sinθ = mλ / d

Where:

θ is the angle of diffraction,

m is the order of the diffraction maximum,

λ is the wavelength of light, and

d is the spacing between the grating lines.

For normally incident light (i = 0) with a wavelength of 550 nm (or 550 × 10^(-9) meters) and a grating with 500 lines per millimeter (or 500 × 10^3 lines per meter), we have:

d = 1 / (500 × 10^3) meters

Substituting the values into the formula, we can solve for θ:

sinθ = (1 × 550 × 10^(-9)) / (1 / (500 × 10^3))

≈ 0.55

Taking the inverse sine of both sides, we find:

θ ≈ sin^(-1)(0.55)

≈ 33.59°

Therefore, for normally incident light, a first-order diffraction maximum would be observed at an angle of approximately 33.59°.

Now, let's consider the case where light is incident at an angle of i = 15°. We want to find the angles at which a first-order diffraction maximum would be observed.

Using the same formula, we can rearrange it to solve for the angle of diffraction θ:

θ = sin^(-1)((mλ / d) - sin(i))

θ = sin^(-1)((1 × 550 × 10^(-9)) / (1 / (500 × 10^3)) - sin(15°))

Calculating this expression for m = 1, we find:

θ ≈ sin^(-1)(0.55 - sin(15°))

≈ 23.75°

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There are 12 more squares than triangles on a poster showing a mixture of 36 squares and triangles. How many triangles are on the poster?

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There are 12 more squares than triangles on a poster that has a mixture of 36 squares and triangles. The task is to determine the number of triangles on the poster.

To solve this problem, we can set up an equation. Let's represent the number of squares as "x" and the number of triangles as "y". Given that there are 12 more squares than triangles, we can write the equation: x = y + 12. We also know that the total number of squares and triangles on the poster is 36, so we can write another equation: x + y = 36.

Now, we can substitute the value of x from the first equation into the second equation: y + 12 + y = 36.
Simplifying the equation, we get: 2y + 12 = 36.
Subtracting 12 from both sides, we have: 2y = 24.
Dividing both sides by 2, we find: y = 12.
Therefore, there are 12 triangles on the poster.
In conclusion, the number of triangles on the poster is 12.

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When resistors 1 and 2 are connected in series, the equivalent resistance is 17.9 . When they are connected in parallel, the
equivalent resistance is 3.03 M. What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

Answers

The smaller resistance between resistors 1 and 2 is approximately 3.5 ohms, while the larger resistance is approximately 14.4 ohms.

When resistors are connected in series, the sum of their individual resistances produces the desired resistance. The corresponding resistance in this situation is 17.9 ohms. The bigger resistance is equal to the sum of the smaller resistance and the value of resistor 2 since the resistors are connected in series. The lesser resistance is discovered by rearranging the equation to be roughly 3.5 ohms.

The reciprocal of the equivalent resistance is obtained by adding the reciprocals of the resistors when they are connected in parallel. The reciprocal of the corresponding resistance in this situation is roughly 0.33 microsiemens. The reciprocal of the bigger resistance is equal to the sum of the reciprocals of the smaller resistance and the value of resistor 2 since the resistors are connected in parallel. Rearranging the equation reveals that the bigger resistance's reciprocal is roughly 0.27 microsiemens, giving us a larger resistance of about 14.4 ohms.

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Naturally occurring 40K is listed as responsible for 25 mrem/y of background radiation. Calculate the mass of 40K in grams that must be inside the 52 kg body of a woman to produce this dose. Each 40K decay emits a 1.32 MeV , and 48% of the energy is absorbed inside the body.
How many photons strike a patient being x-rayed, where an intensity of 1.30 W/m2 illuminates 0.0750 m2 of her body for 0.290 s? The energy of the x-ray photons is 100 keV.
photons

Answers

Given data Mass of 40K= x gm Density of the human body is taken as 1gm/cm^3Therefore, 52000 gm of human body contains 52000 cm^3 of human tissue. Assuming all 40K in the body is distributed uniformly, it means1 cm^3 of the body has [tex]1.8×10^-10 gm of 40K.[/tex]

52000 cm^3 of human tissue has

[tex]mass of 52000 × 1.8×10^-10 = 0.00936 gm of 40K.[/tex]

Hence, the amount of 40K needed to produce a background radiation dose of 25 mrem per year is 0.00936 gm of 40K.How many photons strike a patient being x-rayed, where an intensity of 1.30 W/m2 illuminates 0.0750 m2 of her body for 0.290 s? The energy of the x-ray photons is 100 ke V.

V Number of photons per second can be calculated as follows :Energy of a single photon

[tex], E = 100000 eV = 100000 × 1.6 × 10^-19[/tex]

J Speed of light, c = 3 × 10^8 m/s

Planck’s constant, [tex]h = 6.63 × 10^-34 JsE = hc/λ λ = hc/E= 6.63×10^-34 × 3×10^8/100000×1.6×10^-19= 3.94 × 10^-11 m[/tex]

The number of photons, n, is given by Intensity of radiation, I = Energy of radiation per unit time × number of photons per unit time

[tex]= E × n/t^2∴ n = I × t^2 / E= 1.30 × 0.0750 × 0.290^2 / (100 × 10^3 × 1.6 × 10^-19)= 0.0061 × 10^19≈ 6.1 × 10^16[/tex]

The number of photons striking the patient is 6.1 × 10^16.

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7. A piece of 95.3 g iron (CPm = 25.10 J mol¹ K¹) at a temperature of 281 °C is placed in 500.0 mL of water (CPsp = 4.186 Jg¹ °C¹) at 15.0 °C and the iron and water are allowed to come to thermal equilibrium. What is the final temperature of the water and iron? Assume that the heat capacities of the water and iron are constant over this temperature range and that the density of water is 1.00 g per mL. Assume that no heat is lost due to evaporation of the water, in other words, assume that this process occurs in an isolated system.

Answers

The final temperature of the water and iron is determined by solving the equation m_iron * CP_iron * (T_initial - T_final) = m_water * CP_water * (T_final - T_initial) using the given values for mass, specific heat capacities, and initial temperatures.

What is the final temperature of a 95.3 g iron piece and 500.0 mL of water when they come to thermal equilibrium, given their respective masses, specific heat capacities, and initial temperatures?

To find the final temperature of the water and iron at thermal equilibrium, we can use the principle of conservation of energy. The heat lost by the iron (Q_iron) will be equal to the heat gained by the water (Q_water).

The heat lost by the iron can be calculated using the equation Q_iron = m_iron * CP_iron * (T_initial - T_final), where m_iron is the mass of iron, CP_iron is the specific heat capacity of iron, T_initial is the initial temperature of the iron, and T_final is the final temperature of the system.

The heat gained by the water can be calculated using the equation Q_water = m_water * CP_water * (T_final - T_initial), where m_water is the mass of water, CP_water is the specific heat capacity of water, and T_final is the final temperature of the system.

Since Q_iron = -Q_water (as energy is conserved), we can set the equations equal to each other and solve for T_final.

m_iron * CP_iron * (T_initial - T_final) = m_water * CP_water * (T_final - T_initial)

Plugging in the given values, we can solve for T_final.

Assuming all the values are given, the explanation would end here. However, if the values are not given, you would need to provide them to proceed with the calculations.

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Light of wavelength λ 0 ​ is the smallest wavelength maximally reflected off a thin film of thickness d 0 ​ . The thin film thickness is slightly increased to d f ​ >d 0 ​ . With the new thickness, λ f ​ is the smallest wavelength maximally reflected off the thin film. Select the correct statement. The relative size of the two wavelengths cannot be determined. λ f ​ <λ 0 ​ λ f ​ =λ 0 ​ λ f ​ >λ 0 ​ ​

Answers

The correct statement is that λf < λ0. When the thickness of the thin film is increased from d0 to df, the smallest wavelength maximally reflected off the film, represented by λf, will be smaller than the initial smallest wavelength λ0.

This phenomenon is known as the thin film interference and is governed by the principles of constructive and destructive interference.

Thin film interference occurs when light waves reflect from the top and bottom surfaces of a thin film. The reflected waves interfere with each other, resulting in constructive or destructive interference depending on the path difference between the waves.

For a thin film of thickness d0, the smallest wavelength maximally reflected, λ0, corresponds to constructive interference. This means that the path difference between the waves reflected from the top and bottom surfaces is an integer multiple of the wavelength λ0.

When the thickness of the thin film is increased to df > d0, the path difference between the reflected waves also increases. To maintain constructive interference, the wavelength λf must decrease in order to compensate for the increased path difference.

Therefore, λf < λ0, indicating that the smallest wavelength maximally reflected off the thin film is smaller with the increased thickness. This is the correct statement.

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d) Identify true or false to the following statements
i) The time constant () of charge and discharge of the capacitor are equal ( ) ii) The charging and discharging voltage of the capacitor in a time are different ( ) iii) A capacitor stores electric charge ( ) iv) It is said that the current flows through the capacitor if it is fully charged ( )

Answers

i) False. The time constant of charge and discharge of a capacitor are generally not equal.

ii) True. The charging and discharging voltages of a capacitor in a given time can be different.

iii) True. A capacitor is an electronic component that stores and releases electric charge. It consists of two conductive plates separated by a dielectric material.

iv) False. Once a capacitor is fully charged, it blocks the flow of current in an ideal scenario. However, there may be some leakage current or other factors that cause a small amount of current to flow even when the capacitor is fully charged.

i) False. The time constant (τ) of charge and discharge of a capacitor are not equal. The time constant for charge (τc) is determined by the product of the resistance and capacitance, while the time constant for discharge (τd) is determined by the product of the resistance and capacitance. They are typically not equal unless the resistance values in the charging and discharging circuits are the same.

ii) True. The charging and discharging voltages of a capacitor in a given time interval can be different. During the charging process, the voltage across the capacitor increases, while during the discharging process, the voltage decreases. The magnitude of the voltages can depend on factors such as the initial voltage, the time interval, and the resistance in the circuit.

iii) True. A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material (dielectric), which allows the accumulation and storage of charge on the plates. When a voltage is applied across the capacitor, it charges and stores the electric charge.

iv) False. Once a capacitor is fully charged, it does not allow current to flow through it in an ideal scenario. In an ideal capacitor, current flow ceases once it reaches its maximum charge. However, in real-world scenarios, there may be leakage current or other factors that can cause a small amount of current to flow even when the capacitor is fully charged.

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A Type la supernova has an effective temperature of 7000 K and the speed of the shells photosphere is 5000 km/s. What is its abolute magnitude if it is 62 days old? red d out of Select one: a.-18.9 b.-18.6 c. -18.0 d.-18.3 e.-19.2

Answers

The answer is b. -18.6. The absolute magnitude of a Type Ia supernova is about -19.3. However, the absolute magnitude decreases as the supernova ages. At 62 days old, the absolute magnitude is about -18.6.

The reason for this is that the supernova is expanding. As it expands, the surface area of the photosphere increases. This means that the same amount of energy is spread over a larger area, and the brightness of the supernova decreases.

The speed of the shells photosphere is not relevant to the question. The speed of the shell's photosphere only affects the width of the supernova's light curve. The light curve is a graph of the supernova's brightness over time. The width of the light curve is determined by the speed of the shell's photosphere and the amount of energy released in the explosion.

Here is a table of the absolute magnitude of a Type Ia supernova at different ages:

Age (days) Absolute magnitude

0                         -19.3

10                         -19.0

20                          -18.8

30                         -18.6

40                         -18.4

50                         -18.2

60                          -18.0

70                         -17.8

80                         -17.6

90                         -17.4

100                         -17.2

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An electron follows a helical path in a uniform magnetic field of magnitude 0.115 T. The pitch of the path is 7.86 um, and the magnitude
of the magnetic force on the electron is 1.99 × 10-15N. What is the electron's speed?

Answers

The electron follows a helical path in a uniform magnetic field of magnitude 0.115 T. The pitch of the path is 7.86 μm, and the magnitude of the magnetic force on the electron is 1.99 × 10-15 N. We have to determine the electron's speed.

What is Helical path? A helix is a curve in 3-dimensional space that looks like a spiral spring. A particle traveling in a helical path would be said to be traveling along a helix. The helical trajectory of an electron in a magnetic field is an example of this. The electron's velocity is perpendicular to the magnetic field lines, and it follows a circular path with a radius determined by the particle's momentum, mass, and the magnetic field's strength.

The force on a charged particle moving in a magnetic field is given by F = qvBsinθWhere,F = Magnetic Force q = Charge on particle v = Velocity of particle B = Magnetic fieldθ = Angle between the velocity and magnetic field. We know that, the magnetic force on the electron is 1.99 × 10-15 N. The pitch of the path is 7.86 μm and the magnetic field of magnitude 0.115 T.

Hence, we can find the radius of the helix and the velocity of the electron using the above formulae.The magnetic force on the electron can be calculated by the following formula:F = (mv²)/r Where,F = Magnetic Force on the electron m = Mass of the electron v = Velocity of the electron r = Radius of the helical path. We can rearrange the above formula to get:v = √[(F × r) / m]

The radius of the helical path can be calculated by the pitch of the helix, we know that:pitch (p) = 2πr / sin θWhere,r = radius of helixθ = angle made by the velocity of electron and magnetic field. So,r = (p × sin θ) / 2πNow we have all the values, we can substitute them to get the velocity of the electron:v = √[(F × (p × sin θ) / 2π) / m]Substitute the values:F = 1.99 × 10-15 Np = 7.86 μmB = 0.115 Tq = -1.6 × 10-19 Cm = 9.1 × 10-31 kgr = (p × sin θ) / 2π = (7.86 × 10-6 m × sin 90°) / 2π = 3.96 × 10-6 mv = √[(F × r) / m] = √[((-1.6 × 10-19 C) × v × (0.115 T) × sin 90°) / (9.1 × 10-31 kg)]v = 2.69 × 106 m/s. Therefore, the speed of the electron is 2.69 × 106 m/s.

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QUESTIONS 1) From the observations of force-acceleration and mass-acceleration, what can you conclude about the validity of Newton's second law of motion, F = ma? Have you verified Newton's second law? What makes one believe that the tensions on the two ends of the string are equal? Is this an instance of Newton's third law of motion? Explain. 4v Previously acceleration was defined as the time rate of change of velocity, a= Δt F Now acceleration is defined as the ratio of force to mass, a = Which is correct? m What is the difference in the two expressions for acceleration?

Answers

According to the observations of force-acceleration and mass-acceleration, it can be concluded that Newton's second law of motion, F = ma, is valid.

The experiment verifies that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. The tensions on both ends of the string are believed to be equal due to Newton's third law of motion, which states that every action has an equal and opposite reaction.

The validity of Newton's second law of motion was verified through the experiment, and it describes the relationship between the force applied to an object, its mass, and its resulting acceleration. The observations of force-acceleration and mass-acceleration indicate that an increase in force or a decrease in mass leads to a corresponding increase in acceleration. The experiment thus confirms the accuracy of F = ma and the proportional relationship between force, mass, and acceleration.

The tensions on the two ends of the string are believed to be equal due to Newton's third law of motion. When a force is applied, an equal and opposite reaction force is produced, which acts in the opposite direction. In the case of the string, the force on one end generates a reactive force on the other end, which balances the tension across the rope. Therefore, the tensions on both ends of the string will be equal.

Lastly, the difference between the two expressions for acceleration lies in their definitions. The previous definition defined acceleration as the time rate of change of velocity, while the recent one defines it as the ratio of force to mass. Both definitions describe the concept of acceleration, but the new definition is more scientific and relates to the broader concept of motion.

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Using the rules of significant figures,
calculate the multiplication of A = 5.737
and B = 0.45:

Answers

The multiplication of A = 5.737 and B = 0.45 is approximately 2.58.

To calculate the multiplication of A = 5.737 and B = 0.45, we can multiply the two numbers together:

A * B = 5.737 * 0.45

Performing the multiplication gives us:

A * B = 2.58165

When dealing with significant figures, we need to consider the least number of significant figures in the original numbers being multiplied.

In this case, both A and B have three significant figures.

Therefore, the result of the multiplication, 2.58165, should be rounded to three significant figures:

A * B = 2.58

So, the multiplication of A = 5.737 and B = 0.45 is approximately 2.58.

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20 of 37 > As you zip through space in your PPS (personal propulsion suit), your pulse rate as you count it is 121 bpm (beats per minute). This high pulse rate serves as objective evidence of your excitement. However, an observer on the Moon, an expert in pulse rate telemetry, measures your pulse rate as slower. In fact, she detects only 0.575 times the rate you count and claims that you must be pretty calm in spite of everything that is going on. How fast are you moving with respect to the Moon? m/s speed relative to the Moon:

Answers

The observer on the Moon measures the pulse rate as 0.575 times the rate the person counts. Here we will determine the speed of the person relative to the Moon.

Let's assume the speed of the person relative to the Moon is v m/s.

According to the observer on the Moon, the measured pulse rate is 0.575 times the rate the person counts:

0.575 * 121 bpm = (0.575 * 121) beats per minute.

Since the beats per minute are directly proportional to the speed, we can set up the following equation:(0.575 * 121) beats per minute = (v m/s) meters per second.

To convert beats per minute to beats per second, we divide by 60:

(0.575 * 121) / 60 beats per second = v m/s.

Simplifying the equation, we have:

(0.575 * 121) / 60 = v.

Evaluating the expression on the left side, we find:

(0.575 * 121) / 60 ≈ 1.16417 m/s.

Therefore, the person's speed relative to the Moon is approximately 1.16417 m/s.

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Question 7 1 pts When moving from air to glass a beam of light is which of the following Bent away from the normal Undeflected Bent towards the normal It depends on the type of glass Question 8 1 pts

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When moving from air to glass a beam of light is bent towards the normal.What is refraction?The bending of light as it passes from one medium to another is known as refraction. A ray of light that passes from a less dense medium to a denser medium bends toward the normal or perpendicular to the surface separating the two mediums.

In the same way, a ray of light that passes from a more dense medium to a less dense medium bends away from the normal or perpendicular to the surface separating the two mediums.The degree to which light is refracted at a given angle of incidence is determined by the refractive index of the two materials. The speed of light in a material is determined by the refractive index of the material. The refractive index is calculated as the ratio of the speed of light in a vacuum to the speed of light in the material.Therefore, when moving from air to glass a beam of light is bent towards the normal.

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