Answer:
C. The maximum speed of emitted electrons increase
D. The stopping potential increases
Explanation:
Albert Einstein provided a successful explanation of the photoelectric effect on the basis of quantum theory. He proposed that,
“An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the energy of electrons too.”
Therefore, due to this increase in energy of photons, the kinetic energy of emitted electrons also increase. And the increase in kinetic energy follows with the increase in velocity.
Now, the stopping potential is directly proportional to Kinetic Energy of the electrons. Thus, the increase in Kinetic Energy, results in an increase in stopping potential.
Therefore, two options apply here:
C. The maximum speed of emitted electrons increases
D. The stopping potential increases
If the frequency of the incident light in the photoelectric effect is increased, C. The maximum speed of the emitted electrons increases and D. The stopping potential increases; however, the work function does not change and the number of electrons emitted per second remains the same at a constant light intensity.
For the question about what happens if the frequency of the incident light in the photoelectric effect is increased, the correct options are:
The maximum speed of the emitted electrons increases.The stopping potential increases.The work function of the metal (Option A) is a material property and does not change with the frequency of the incident light. Therefore, this option is incorrect. The number of electrons emitted from the metal per second (Option B) depends on the intensity of the light, not its frequency, so increasing the frequency at constant intensity does not increase the emission rate. Thus, this option is also incorrect. When the frequency of the incident light is increased, the kinetic energy of the emitted electrons increases (as per Einstein's equation for the photoelectric effect), which correlates with Option C. The stopping potential, which is the electric potential needed to stop the fastest photoelectrons, also increases with the electron's kinetic energy, confirming Option D as correct.
Double the resistance of the resistor by changing it from 10 Ω to 20 Ω. What happens to the current flowing through the circuit?
Answer:
2 amps
Explanation:
you just divide
The power liens that run through your neighborhood carry alternating currents that reverse direction 120 times per second. As current changes so does magnetic field. If you put a loop of wire up near the power line to extract power by tapping the magnetic field, sketch how you would orient the coil of wire next to a power line to develop the max emf in the coilo If magnetic flux through a loop changes, induced emf is produced
o If the area of the loop is parallel to the field, the flux through the loop is minimum
• So no emf is produced
o To get the most of the flux through the loop you place it closer to the power lines and in orientations so the plane of the loop also contains power lines
• Flux would be max and result in greater induced emf as field goes from max to zero to max and then in other direction
Answer:
the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
Explanation:
Faraday's law is
ε = - dΦ / dt
where Ф magnetic flow
the flow is
Ф = B. dA = B dA cos θ
therefore, to obtain the maximum energy, the cosine function must be maximum, for this the direction of the fluctuating magnetic field and the normal direction to the area must be parallel.
The magnetic field in a cable through which current flows is circular, so the loop must be perpendicular to the wire, this is the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
an outline is shown in the attachment
the correct answer is b
A guitar string vibrates at a frequency of 330Hz with wavelength 1.40m. The frequency and wavelength of this sound wave in air (20 C) when it reaches our ears is: Lower frequency, same wavelength Same frequency, same wavelength Higher frequency, same wavelength Same frequency, shorter wavelength
In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.
Speed of wave:here the speed of the wave should be provided by the following equation.
v = fλ
λ = v/f
here,
f = Frequency
λ = Wavelength
Also, the wavelength is directly proportional to the velocity.
So, the frequency remains same and the wavelength shortens.
So based on this we can say that In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.
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Final answer:
The sound wave produced by a vibrating guitar string will maintain the same frequency as it travels through the air to our ears because the frequency of a wave depends on its source and doesn't change in a uniform medium. Although the wavelength can change in different mediums, the question context does not specify a medium change; hence, the frequency and wavelength when heard would be the same as produced by the string.
Explanation:
The question is about the behavior of sound waves produced by a vibrating guitar string when they travel through the air. Specifically, it examines how the frequency and wavelength of the sound wave change once it reaches the listener's ear compared to when it is produced by the string.
It's important to note that when a guitar string vibrates at a certain frequency, it creates sound waves that propagate through the air with the same frequency. This is because the frequency of a wave is determined by its source and remains constant as it travels through a uniform medium. However, the wavelength may change if the speed of the wave changes, which depends on the medium's properties.
In the context of sound waves in air at 20°C, the speed of sound is approximately 343 m/s. Considering this, the frequency of the sound wave when it reaches our ears remains the same as the frequency of the vibrating string, because the frequency of a wave only changes if the speed of the wave's source relative to the observer changes, which is not the case here.
The wavelength of the sound wave in the air might differ from the wavelength on the string due to the different speeds of wave propagation in different mediums (the string versus air), but for the purpose of this specific question, such details are beyond the scope provided.
Therefore, the correct answer is 'Same frequency, same wavelength', under the assumption that both the measurements were taken in the air or that the provided wavelength already accounts for the transition from the string to the air.
1. (4 pts) We use Fst to measure the number of migrants per generation (Nm) by assuming a balance between the effects of a differentiating evolutionary force and a cohesive evolutionary force. Name those two forces. (Hint, think of the computer simulations of migration)
Answer:
This structure is determined by the combined effect of deterministic forces and stochastic forces.
Explanation:
Genetic drift refers to random fluctuations in allele frequencies due to chance events , a stochastic (random) force which scrambles the predictable effects of selection, mutation, and gene flow.
Genetic drift is not a potent evolutionary force in very large randomly mating populations. to illustrate the consequences of genetic drift we will consider what happens when drift alone is altering the frequencies of alleles among many small populations. To prove this, we need to understand Population structure, which describes how individuals (or allele frequencies) in breeding populations vary in time and space.
Construct a parallel-plate capacitor where a second line of charges equal in size and opposite in charge are placed below the line of positive charges. Examine what the "E-field" is like between the plates using a sensor.
A parallel-plate capacitor consists of two plates with opposite charges, creating an electric field between them. The electric field can be measured using an electric field sensor.
Explanation:In a parallel-plate capacitor, a second line of charges equal in size but opposite in charge is placed below the line of positive charges, creating an electric field (E-field) between the plates. The E-field is a vector quantity that points from the positive plate to the negative plate. Its magnitude can be determined using a sensor, such as an electric field sensor, which measures the strength of the electric field.
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A parallel-plate capacitor creates a uniform electric field between its plates, where the field's magnitude is proportional to the surface charge density and directly proportional to the amount of charge on the plates.
Explanation:A parallel-plate capacitor consists of two identical, parallel conducting plates separated by a certain distance. When we create a system where one plate is charged with positive charges and the other with an equal magnitude of negative charges, by, for instance, connecting the plates to the opposite terminals of a battery, an electric field (often referred to as the E-field) is established between the plates. This electric field is extremely uniform because of the geometry of the plates and the even distribution of charges.
The magnitude of the electric field (E) between the plates is proportional to the surface charge density (σ), which is the charge per unit area on a plate. It is described mathematically as E = σ/ε0, where ε0 is the permittivity of free space. The strength of the electric field is also directly proportional to the amount of charge (Q) on the plates. Therefore, when more charge is present on the plates, the electric field will be stronger, illustrated by the increase in the density of field lines that begin on positive charges and end on negative ones.
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The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in this glass strikes the glass–air interface at a 25.50∘ angle of incidence and refracts out into the air. What is the angular separation between the red and violet components of the spectrum that emerges from the glass?
Answer:
10.16 degrees
Explanation:
Apply Snells Law for both wavelenghts
\(n_{1}sin\theta_{1} = n_{2}sin\theta_{2}\)
For red
(1.620)(sin 25.5) = (1)(sin r)
For red, the angle is 35.45degrees
For violet
(1.660)(sin 25.5) = (1)(sin v)
For violet, the angle is 45.6 degrees
The difference is 45.6- 35.45 = 10.16 degrees
Which of the following is NOT a cause for the intervention of antitrust authorities?
a.
Companies abuse their market power by acquiring new firms that allow for the increase of prices above a level that would occur in a competitive market.
b.
The merger between two companies allows for several customer options and substantial competition within the industry.
c.
An acquisition that creates too much consolidation and increases the potential for future abuse of market power.
d.
A company cuts prices when a new competitor enters the industry to force the competitor out of business.
e.
Dominant companies use their market power to crush potential competitors.
Answer:
b.
The merger between two companies allows for several customer options and substantial competition within the industry.
Explanation:
Antitrust authorities will only come in if the merger was to remove competition and reduce consumer options.
Final answer:
Antitrust authorities intervene when firms engage in practices that reduce competition. These practices include abusive use of market power, anticompetitive mergers and acquisitions, and restrictive practices. A merger enhancing competition does NOT warrant such intervention. Thus, option B is correct.
Explanation:
The intervention of antitrust authorities is generally warranted when actions are taken by firms that reduce competition and harm the economic ideal of a competitive market. These actions may include abusive use of market power, mergers and acquisitions that significantly increase market concentration and reduce competition, and various restrictive practices that limit competition, such as tie-in sales, bundling, and predatory pricing.
However, not all actions that affect competition trigger antitrust intervention. For example, if the merger between two companies allows for several customer options and substantial competition within the industry, that would not be a cause for concern for antitrust authorities. In this case, the merger does not lead to abusive market power or reduce competition, and therefore, option b 'The merger between two companies allows for several customer options and substantial competition within the industry' would be the correct answer as it is NOT a cause for intervention by antitrust authorities.
A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times before coming to rest. Find the constant angular acceleration of the centrifuge (in rad/s2). Consider the direction of the initial angular velocity to be the positive direction, and include the appropriate sign in your result.
Answer:
Explanation:
Given,
initial angular speed, ω = 3,700 rev/min
= [tex]3700\times \dfrac{2\pi}{60}=387.27\ rad/s[/tex]
final angular speed = 0 rad/s
Number of time it rotates= 46 times
angular displacement, θ = 2π x 46 = 92 π
Angular acceleration
[tex]\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}[/tex]
[tex]\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}[/tex]
[tex]\alpha = -259.28 rad/s^2[/tex]
A pendulum string with the length of 105 cm has the 325 g bob attached to it. At the lowest point of the swing the pendulum bob moves with the speed of 2.15 m/s. Determine: a) the centripetal acceleration of the pendulum bob at this point; b) the force of tension in the string at this point; c) the kinetic energy of the pendulum bob at this point;
Answer:
A. 4.40m/s²
B. 4.615N
C. 0.7511J
Explanation:
A. Ac= V²/L
= (2.15)²/1.05
= 4.50m/s²
B. T =Mv²/L + mg
0.325(4.40)+ 0.325(9.8)
= 4.615N
C. K.E = 1/2mv²
=0.7511J
You can also refer to attached handwritten document for more details
Answer:
(a) 4.4 m/s²
(b) 4.615 N.
(c) 0.751 J
Explanation:
(a)
Using,
a' = v²/r..................... Equation 1
Where a' = centripetal acceleration, v = speed of the pendulum bob, r = radius or length of the pendulum bob.
Given: v = 2.15 m/s, r = 105 cm = 1.05 m
Substitute into equation 1
a' = 2.15²/1.05
a' = 4.4 m/s².
(b) The force of tension in the string = Tangential force + weight of the bob.
T = ma'+mg..................... Equation 2
Where T = Force of tension in the string, m = mass of the bob, g = acceleration due to gravity.
Given: m = 325 g = 0.325 kg, a' = 4.4 m/s², g = 9.8 m/s²
Substitute into equation 2
T = 0.325×4.4+0.3325×9.8
T = 1.43+3.185
T = 4.615 N.
(c) Kinetic energy of he bob at that point = 1/2mv²
Ek = 1/2mv²...................... Equation 3
Where Ek = kinetic energy of the bob
Given: m = 0.325 kg, v = 2.15 m/s
Substitute into equation 3
Ek = 1/2(0.325)(2.15²)
Ek = 0.751 J
The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2 mm^2 , its average length is 26 cm , and its average Young's modulus is 1474 MPa . How much tensile stress is required to stretch this muscle by 5.2% of its length? If we model the tendon as a spring, what is its force constant? If a 75 kg sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch?
Answer:
1) tensile stress = 76.648 Mpa
2) extension = 0.0215 m
Explanation:
Detailed explanation and calculation is shown in the image below
Final answer:
Tensile stress and the force constant for the Achilles tendon can be calculated using Young's modulus, Hooke's Law, and the given dimensions. A 75 kg sprinter exerting a force 13 times his weight would cause a specific amount of stretch in the tendon, calculated by applying Hooke's Law with the determined spring constant.
Explanation:
The tensile stress required to stretch the Achilles tendon by 5.2% of its length can be calculated using Hooke's Law and the definition of stress and strain. Stress (σ) is the force applied per unit area (A) and strain (ε) is the relative change in length (ΔL/L). The Young's modulus (E) relates stress and strain through σ = Eε.
First, calculate the strain which is 5.2% or 0.052. Since the Young's modulus (E) is given as 1474 MPa and the strain (ε) is 0.052, we can now calculate the stress:
σ = Eε = 1474 MPa × 0.052 = 76.648 MPa
Next, we model the tendon as a spring to find the force constant (k). Hooke's Law states that F = kΔx, where F is the force applied and Δx is the change in length. To find k, we rearrange the formula to k = F/Δx. Here, F is the tensile force, which is σ × A. We substitute the stress we found and the area of the tendon to find k.
For a 75 kg sprinter exerting a force of 13.0 times his weight, the force F would be 13.0 × 75 kg × 9.8 m/s² (acceleration due to gravity). To find the stretch (Δx), we use Hooke's Law again with the calculated force constant k.
The Achilles tendon will stretch according to that force, proportionally to the spring constant k. The exact calculations require substitution of the calculated force into the Hooke's Law equation to solve for Δx.
Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheets. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2?
Answer:
Explanation:
solution found below
The electric field at a midpoint between two finite charged sheets can be calculated using Gauss's Law. While the field strength would theoretically be zero if the sheets were infinite, because real sheets are finite, the field will not be exactly zero.
Explanation:This question concerns the physics concept of electric fields originating from two finite charged sheets. Given a point 'a/2' on the x-axis in-between these sheets, the magnitude E of the electric field can be calculated using Gauss's Law. The electric field E due to an infinite sheet of charge is given by σ/2ε₀ (σ is the charge density), which is independent of the distance from the sheet. Hence, for a point which is midway between two such sheets, and each sheet carries charge of opposite polarity, the field strength at that point will be zero. However, since real sheets will not be infinitely large, the field will not be exactly zero. The exact value will depend on the exact geometry of the problem and on the distance from the sheets to the point.
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Pease circle the statements incompatible with the Kelvin-Planck Statement. (A) No heat engine can have a thermal efficiency of 100%. (B) It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work without rejecting waste heat to a cool reservoir. (C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available. (D) Any device that violates the Kelvin-Planck statement also violates the Clausius statement, and vice versa.
Answer:
(C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available.
Explanation:
The above option was never stated in the law
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2, where xo and yo are measured in feet and t is measured in seconds. The angular displacement of a radial line measured from a vertical reference line is θ = 8t4, where θ is measured in radians. Determine the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s.
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
the peripheral velocity that is directed downward [tex](-V_y)[/tex] along the y-axisthe linear velocity [tex](V_x)[/tex] that is directed along the x-axisNow;
[tex]V_x = \frac{d}{dt}(12t^3+2) = 36 t^2[/tex]
[tex]V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s[/tex]
Also,
[tex]-V_y = R* \omega[/tex]
where [tex]\omega[/tex](angular velocity) = [tex]\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)[/tex]
[tex]-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s[/tex]
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
the cart. The coefficient of kinetic friction between the person and the cart is 0.480. Friction between the cart and ground can be ignored. (Let the positive direction be to the right.) (a) Find the final velocity of the person and cart relative to the ground. (Indicate the direction with the sign of your answer.) m/s (b) Find the frictional force acting on the person while he is sliding across the top surface of the cart. (Indicate the direction with the sign of your answer.) N (c) How long does the frictional force act on the person? s (d) Find the change in momentum of the person. (Indicate the direction with the sign of your answer.) N · s Find the change in momentum of the cart. (Indicate the direction with the sign of your answer.) N · s (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (Indicate the direction with the sign of your answer.) m (f) Determine the displacement of the cart relative to the ground while the person is sliding. (Indicate the direction with the sign of your answer.) m (g) Find the change in kinetic energy of the person. J (h) Find the change in kinetic energy of the cart. J (i) Explain why the answers to (g) and (h) differ. (What kind of collision is this one, and what accounts for the loss of mechanical energy?)
Answer:
(a) Find the final velocity of the person and cart relative to the ground. = 1.367m/s
(b) Find the frictional force acting on the person while he is sliding across the top surface of the cart = 286.944N
(c) How long does the frictional force act on the person? = 0.5809s
(d) Find the change in momentum of the person. = 166.774N
(e) Determine the displacement of the person relative to the ground while he is sliding on the cart. = 1.5878m
(f) Determine the displacement of the cart relative to the ground while the person is sliding. = 0.3970m
(g) Find the change in kinetic energy of the person. = -455.71J
(h) Find the change in kinetic energy of the cart = 113.99j
(i) Explain why the answers to (g) and (h) differ.
the force exerted on the cart by the person must be equal in magnitude and opposite in direction to the force exerted on the cart by the person. The changes in momentum of the two cart must be equal in magnitude and must add up to zero. The following represents two way thinking about "WAY". The distance the cart moves is different from the distance moved by the point of application of frictional force on the cart.
Explanation:
check attachment for explanation.
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released
Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 260000 kg and a velocity of 0.32 m/s in the horizontal direction, and the second having a mass of 52500 kg and a velocity of -0.14 m/s in the horizontal direction. What is their final velocity?
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s
The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0 Ω⋅m. The conducting path between the hands can be represented approximately as a cylinder 1.6 m long and 0.10 m in diameter. The skin resistance can be made negligible by soaking the hands in salt water (a) What is the resistance between the hands if the skin resistance is negligible? (b) if skin resistance is negligible, what potential difference between the hands is needed for lethal shock current 100 mA?
Answer:
A) R = 1019.11 Ω
B) V = 101.91 V
Explanation:
a) The resistance is given by the formula;
R = ρL/A
Where;
ρ is bulk density = 5.0 Ωm
L is length of cylinder = 1.6m
A is area = (π/4)D² = (π/4) x 0.1² = 0.007854 m²
Plugging in the relevant values, we obtain ;
R = (5 Ωm x 1.6m)/(0.007854 m²)
R = 1019.11 Ω
b) potential difference is given by;
V = RI
Thus, plugging in relevant values, we obtain;
V = 1019.11 Ω ∙ 0.1A
V = 101.91 V
The resistance between the hands if the skin resistance is negligible is approximately [tex]\( 1019 \, \Omega \)[/tex], the potential difference between the hands needed for a lethal shock current of 100 mA is approximately [tex]\( 101.9 \, \text{V} \).[/tex]
To solve this problem, we'll use the formulas for resistivity and Ohm's law.
Part (a): Resistance Between the Hands
The resistance R of a cylindrical conductor can be calculated using the formula:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
where:
- [tex]\( \rho \)[/tex] is the resistivity of the material (in this case, the human body's resistivity is [tex](\( 5.0 \, \Omega \cdot \text{m} \))[/tex].
- L is the length of the conductor (1.6 m).
- A is the cross-sectional area of the conductor.
First, we need to calculate the cross-sectional area A of the cylindrical conductor. The cross-sectional area of a cylinder is given by:
[tex]\[ A = \pi \left(\frac{d}{2}\right)^2 \][/tex]
where d is the diameter of the cylinder. Here, [tex]\( d = 0.10 \, \text{m} \)[/tex].
So,
[tex]\[ A = \pi \left(\frac{0.10 \, \text{m}}{2}\right)^2 = \pi \left(0.05 \, \text{m}\right)^2 \]\[ A = \pi \cdot (0.05)^2 \, \text{m}^2 \]\[ A \approx 3.14 \cdot 0.0025 \, \text{m}^2 \]\[ A \approx 0.00785 \, \text{m}^2 \][/tex]
Now we can calculate the resistance R:
[tex]\[ R = 5.0 \, \Omega \cdot \text{m} \frac{1.6 \, \text{m}}{0.00785 \, \text{m}^2} \]\[ R = 5.0 \cdot \frac{1.6}{0.00785} \, \Omega \]\[ R = 5.0 \cdot 203.82 \, \Omega \]\[ R \approx 1019.1 \, \Omega \][/tex]
Part (b): Potential Difference for Lethal Shock Current
Ohm's law states that the voltage V across a resistor is given by:
[tex]\[ V = I \cdot R \][/tex]
where I is the current and R is the resistance.
Given that the lethal shock current is [tex]\( 100 \, \text{mA} = 0.1 \, \text{A} \)[/tex] and the resistance [tex]\( R \) is \( 1019 \, \Omega \)[/tex]:
[tex]\[ V = 0.1 \, \text{A} \cdot 1019 \, \Omega \]\[ V = 101.9 \, \text{V} \][/tex]
A merry-go-round of radius 2.74 m and a moment of inertia of 340 kgm2 rotates without friction. It makes 1 revolution every 4.00 s. A child of mass 25.0 kg sitting at the center crawls out to the rim. Find (a) the new angular speed of the merry-go-round, and (b) the kinetic energy change during this process.
Answer:
a) [tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex], b) [tex]\Delta K = - 149.352\,J[/tex]
Explanation:
a) The initial angular speed of the merry-go-round is:
[tex]\omega_{o} = \left(\frac{1\,rev}{4\,s}\right)\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)[/tex]
[tex]\omega_{o} \approx 1.571\,\frac{rad}{s}[/tex]
The final angular speed of the merry-go-round is computed with the help of the Principle of Angular Momentum:
[tex]\left(340\,\frac{kg}{m^{2}}\right)\cdot \left(1.571\,\frac{rad}{s} \right) = \left[340\,\frac{kg}{m^{2}}+(25\,kg)\cdot (2.74\,m)^{2} \right]\cdot \omega_{f}[/tex]
[tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex]
b) The change in kinetic energy is:
[tex]\Delta K = \frac{1}{2}\cdot \left\{\left[340\,\frac{kg}{m^{2}} + \left(25\,kg\right)\cdot \left(2.74\,m\right)^{2}\right]\cdot \left(1.012\,\frac{rad}{s} \right)^{2} - \left(340\,\frac{kg}{m^{2}} \right)\cdot \left(1.571\,\frac{rad}{s} \right)^{2} \right\}[/tex][tex]\Delta K = - 149.352\,J[/tex]
a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a distance d from the origin. The value of d for which the x component of the force on the second particle is the greatest is
Answer:
Explanation:
Force on Q due to q₁
= k Q.q₁ / r² , r is distance between two.
X component
= k Q.q₁ / r² x cos θ , θ is angle r makes with x - axis.
= k Q.q₁ / r² x cos θ
r = a / sinθ
X component
= k Q.q₁ sin²θ . cosθ/ a²
= k Q.q₁ / 2 a² x sin2θ . sinθ
for maximum value of it ,
sin2θ = 1 , θ = 45°
a / d = tan45
a = d .
A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, the light takes 1.82 ns longer to travel its length. What is the refractive index of this jelly?
Answer:
1.208
Explanation:
L = Length of tube
c = Speed of light in air
v = Speed of light in jelly
Time taken by light in tube
[tex]\dfrac{L}{c}=8.72[/tex]
Time taken when jelly is present
[tex]\dfrac{L}{v}=8.72+1.82\\\Rightarrow \dfrac{L}{v}=10.54\ ns[/tex]
Dividing the above equations we get
[tex]\dfrac{v}{c}=\dfrac{8.72}{10.54}\\\Rightarrow \dfrac{c}{v}=\dfrac{10.54}{8.72}\\\Rightarrow n=1.208[/tex]
The refractive index of the jelly is 1.208
Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia 3.76 kg·m2 about its central axis, is set spinning counterclockwise (which may be taken as the positive direction) at 436 rev/min. The second disk, with rotational inertia 9.20 kg·m2 about its central axis, is set spinning counterclockwise at 953 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at 953 rev/min, what are their (b) angular velocity (using the correct sign for direction) and (c) direction of rotation after they couple together?
Answer:
A) ω = 13.38 rev/s
B) ω = 9.167 rev/s
C) In clockwise direction
Explanation:
We are given;
Rotational Inertia of first disk; I_1 = 3.76 kg·m²
Angular velocity of first disk; ω_1 = 436 rev/min = 7.267 rev/s
Rotational Inertia of second disk; I_2 = 9.2 kg·m²
Angular velocity of second disk; ω_2 = 953 rev/min = 15.883 rev/s
Total rotational inertia is;
I_total = I_1 + I_2
I_total = 3.76 + 9.2 = 12.96 kg·m²
Now; total angular momentum will be;
L_total = L_1 + L_2
Where L_1 is angular momentum of first disk and L_2 is angular momentum of second disk.
Thus;
I_total•ω = I_1•ω_1 + I_2•ω_2
Plugging relevant values in, we can find their angular speed after coupling which is ω.
Thus;
12.96ω = (3.76 x 7.267) + (9.2 x 15.883)
12.96ω = 173.44752
ω = 173.44752/12.96
ω = 13.38 rev/s
B) since second disk is now spinning clockwise, thus;
I_total•ω = I_1•ω_1 - I_2•ω_2
12.96ω = (3.76 x 7.267) - (9.2 x 15.883)
12.96ω = -118.8
ω = -118.8/12.96
ω = -9.167 rev/s
The negative sign tells us that it is clockwise.
So we would say ω = 9.167 rev/s in clockwise direction
A spaceship from a friendly, extragalactic planet flies toward Earth at 0.203 0.203 times the speed of light and shines a powerful laser beam toward Earth to signal its approach. The emitted wavelength of the laser light is 691 nm . 691 nm. Find the light's observed wavelength on Earth.
Answer:
567.321nm
Explanation:
See attached handwritten document for more details
Answer:
The observed wavelenght on earth will be 5.51x10^-7 m
Explanation:
Speed of light c = 3x10^8 m/s
Ship speed u = 0.203 x 3x10^8 = 60900000 = 6.09x10^7 m/s
Wavelenght of laser w = 691x10^-9 m
Observed wavelenght w' = w(1 - u/c)
u/c = 0.203
w' = 691x10^-9(1 - 0.203)
w' = 5.51x10^-7 m
Consider a perfectly reflecting mirror oriented so that solar radiation of intensity I is incident upon, and perpendicular to, the reflective surface of the mirror.
(a) If the mirror has surface area A, what is Frad, the magnitude of the average force due to the radiation pressure of the sunlight on the mirror?
Express your answer in terms of the intensity I, the mirror's surface area A, and the speed of light c.
Answer:
Frad = 2IA/C
Explanation:
see attached file
Explicitly solve the Heisenberg equations of motion to find the time–dependent raising and lowering (creation and annihilation) operators for a one-dimensional oscillator in the Heisenberg picture. Show these operators are consistent with the time–dependent position and momentum operators previously derived in Lecture
Answer:
see detailed solution attached.
Explanation:
Final answer:
In the Heisenberg picture of quantum mechanics, the creation and annihilation operators for a one-dimensional harmonic oscillator evolve in time as â(t) = â(0)e-iwt and â¹(t) = â¹(0)eiwt, respectively, which are consistent with the time evolution of the position and momentum operators derived from the Heisenberg equations of motion.
Explanation:
Time-Dependent Creation and Annihilation Operators
In the Heisenberg picture of quantum mechanics, unlike the Schrödinger picture, the state vectors are stationary and the operators evolve with time. For the one-dimensional harmonic oscillator, the time-dependent creation (â¹(t)) and annihilation (â(t)) operators can be solved explicitly. The Heisenberg equations of motion for these operators are given by:
â(t)= â(0)e-iwt
â¹(t) = â¹(0)eiwt
where â(0) and â¹(0) are the initial operators at t=0, and w is the angular frequency of the oscillator. These time-dependent operators are then consistent with the previously derived time-dependent position (q(t)) and momentum (p(t)) operators:
q(t) = q(0)cos(wt) + (p(0)/mw)sin(wt)
p(t) = p(0)cos(wt) - mwq(0)sin(wt)
Here, m is the mass of the oscillator, and w is again the angular frequency. This consistency is because the position and momentum operators can be expressed in terms of the creation and annihilation operators. The expressions for â(t) and â¹(t) in the Heisenberg equations demonstrate that these operators oscillate with time in a manner that corresponds to the harmonic nature of the oscillator's motion in quantum mechanics.
The time evolution of any quantum mechanical operator in the Heisenberg picture can be described by the time evolution operator, U, which is unitary. This allows time propagation of operators while keeping the wavefunction unchanged.
The objective lens of a certain refracting telescope has a diameter of 63.5 cm. The telescope is mounted in a satellite that orbits the Earth at an altitude of 265 km to view objects on the Earth's surface. Assuming an average wavelength of 500 nm, find the minimum distance between two objects on the ground if their images are to be resolved by this lens.
Answer:
0.255 m
Explanation:
We are given that
Diameter=d=63.5 cm=[tex]63.5\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
L=265 km =[tex]265\times 1000=265000 m[/tex]
Wavelength,[tex]\lambda=500nm=500\times 10^{-9} m[/tex]
[tex]1nm=10^{-9} m[/tex]
We have to find the minimum distance between two objects on the ground if their images are to be resolved by this lens.
[tex]sin\theta=1.22\frac{\lambda}{d}[/tex]
[tex]sin\theta=\frac{1.22\times 500\times 10^{-9}}{63.5\times 10^{-2}}[/tex]
[tex]sin\theta=\approx \theta=9.606\times 10^{-7} rad[/tex]
[tex]\frac{y}{L}=tan\theta\approx \theta[/tex]
[tex]y=L\theta=265000\times 9.606\times 10^{-7}=0.255 m[/tex]
A rectangular neoprene sheet has width W = 1.00 m and length L = 4.00 m. The two shorter edges are affixed to rigid steel bars that are used to stretch the sheet taut and horizontal. The force applied to either end of the sheet is F = 81.0 N. The sheet has a total mass M = 4.00 kg. The left edge of the sheet is wiggled vertically in a uniform sinusoidal motion with amplitude A = 10.0 cm and frequency f = 1.00 Hz. This sends waves spanning the width of the sheet rippling from left to right. The right side of the sheet moves upward and downward freely as these waves complete their traversal.
(A) Derive an expression for the velocity with which the waves move along the sheet. Express your answer in terms of the variables F, L, and M.
(B) What is the value of this speed for the specified choices of these parameters? Express your answer with the appropriate units.
Answer:
(a) [tex]\sqrt{FL/M}[/tex]
(b) [tex]9 ms^{-1}[/tex]
Explanation:
(a)
The speed of the wave depends on the type of the material. Here we have neoprene sheet so the speed of the wave will depend on the linear density of neoprene sheet. Linear Density is defined as Mass per unit length of the material (as materials of same type can have different thickness). The symbol used for the Linear Density is .
μ = Mass of the sheet / Length of the sheet
The speed of the wave in such material can be be found by using the equation:
[tex]\frac{1}{v^{2} } = {\frac{Linear Density}{Force} }[/tex] (where v is speed)
The equation can be rearranged:
v = [tex]\sqrt{Force/Linear Density}[/tex]
so,
v = sqrt(F/μ)
v = sqrt(F/ (M/L))
v = [tex]\sqrt{FL/M}[/tex] (answer)
(b)
Putting the values
F = 81 N
M = 4kg
L = 4m
v = [tex]\sqrt{FL/M}[/tex]
v = 9m/s
what is the formula for braking force?
Answer:
Explanation:
The braking force in the context of stopping a vehicle involves the frictional force exerted by the brakes, and it's related to the mass and deceleration of the vehicle. Calculations often use Newton's second law, F = ma, for force, or the work-energy principle, W = F * d * cos(θ), to determine stopping distances.
The formula for braking force isn't provided with a single standard equation, as it relates to several physical quantities, such as friction, mass, and acceleration. However, in the context of vehicle braking, we often analyze the frictional force exerted by the brakes on the car's wheels to determine stopping distance or to calculate the deceleration of the vehicle. The basic physics equation used in this context is Newton's second law, F = ma, where F is the force, m is the mass of the vehicle, and a is the acceleration (which will be negative when braking).
Based on the provided contexts, if a car has a mass (m) and applies a braking force (F), and you want to find the stopping distance (d), you can also use energy equations. The work done by the brakes (W) is equal to the change in the car's kinetic energy, which can be calculated using the equation W = F * d * cos(θ), where θ is the angle at which the force is applied - typically 0 degrees, as the force is in the opposite direction of motion.
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750kg car traveling to the right at 1.60m/s collides with a 1450kg car going to the left at 1.10m/s . Measurements show that the heavier car's speed just after the collision was 0.260m/s in its original direction. You can ignore any road friction during the collision.
A:What was the speed of the lighter car just after the collision?
B:Calculate the change in the combined kinetic energy of the two-car system during this collision.
Answer:
Explanation:
We shall apply law of conservation of momentum .
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ , m₁ ,m₂ are masses of two bodies colliding with velocities u₁ and u₂ respectively . v₁ and v₂ are their velocities after collision.
m₁ = 1750 , m₂ = 1450 , u₁ = 1.6 m /s , u₂ = - 1.1 m /s , v₁ = .26 m /s
substituting the values
1750 x 1.6 + 1450 x - 1.1 = 1750 x .26 + 1450 v₂
2800 - 1595 = 455 + 1450v₂
1450v₂ = 750
v₂ = .517 m /s
B ) initial kinetic energy
= 1/2 x 1750 x 1.6²+ 1/2 x 1450 x -1.1²
= 2240+ 877.25
= 3117.25 J
final kinetic energy
= 1/2 x 1750 x .26²+ 1/2 x 1450 x .517²
= 59.15 + 193.78
= 252.93
loss of kinetic energy
= 3117.25 - 252.93
= 2864.32 J
We will use a video to analyze the dependence of the magnitude of the Coulomb force between two electrically-charged spheres on the distance between the centers of the spheres. The electrical interaction is one of the fundamental forces of nature and acts between any pair of charged objects, therefore it is important to understand how precisely the separation distance affects the corresponding force between them.
1. Study conceptually the nature of Hint 1. Unit conversions Note that 1 nC 109 c. Hint2. The values for the first data point The first suggested data point was 31.9 cm and 0.00016 N electric charge and force
2. Take measurements of the force exerted between two electrically- charged spheres as the distance between them is varied
3. Determine graphically the relationship between electric force and distance nc
Find the attached photo for the solution
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. Initially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 125 ms
what is the average induced emf around the border of the circular region?
Answer:
The average induced emf around the border of the circular region is [tex]8.48\times 10^{-5}\ V[/tex].
Explanation:
Given that,
Radius of circular region, r = 1.5 mm
Initial magnetic field, B = 0
Final magnetic field, B' = 1.5 T
The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V[/tex]
So, the average induced emf around the border of the circular region is [tex]8.48\times 10^{-5}\ V[/tex].
Answer:
[tex]84.8\times 10^{-6} V[/tex]
Explanation:
We are given that
Radius,r=1.5 mm=[tex]1.5\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Initial magnetic field,[tex]B_0=0[/tex]
Final magnetic field,[tex]B=1.5 T[/tex]
Time,[tex]\Delta t=[/tex]125 ms=[tex]125\times 10^{-3} s[/tex]
[tex]1 ms=10^{-3}s [/tex]
We know that average induced emf
[tex]E=\frac{d\phi}{dt}=-\frac{d(BA)}{dt}=A\frac{dB}{dt}=A\frac{(B-B_0)}{dt}[/tex]
Substitute the values
[tex]E_{avg}=\pi (1.5\times 10^{-3})^2\times \frac{1.5-0}{125\times 10^{-3}}[/tex]
[tex]E_{avg}=84.8\times 10^{-6} V[/tex]