When monochromatic light is incident on a metal surface, electrons may be ejected from the surface due to the photoelectric effect.
The energy of the ejected electrons is determined by the energy of the incident photons minus the energy required to overcome the work function of the metal (i.e., the minimum energy required to remove an electron from the metal).
The maximum energy of the ejected electrons is reached when the energy of the incident photons is equal to or greater than the work function of the metal.
If the energy of the incident photons is increased beyond this threshold, the maximum energy of the ejected electrons will not change.
However, if the intensity of the incident light is increased (i.e., the number of photons incident on the metal surface per unit time is increased), the number of ejected electrons per unit time will increase.
Since the energy of each ejected electron is determined by the energy of the incident photons, increasing the number of photons incident on the metal surface per unit time will increase the number of ejected electrons with higher energy (i.e., the maximum energy of the ejected electrons will increase).
This is because more electrons will be able to absorb photons with energies greater than the work function of the metal and gain higher energies as a result.
Therefore, the correct answer is: the maximum energy will increase.
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in the derivation of the wave equation, a simplifying approximation is made. what is it? select answer from the options below the maximum displacement is small. the wave speed is small. newton's second law is valid. the wave shape is sinusoidal. the wave frequency is small.
The simplifying approximation made in the derivation of the wave equation is that the maximum displacement is small.
What is Wave?
A wave is a disturbance that propagates through space or a medium, transporting energy from one point to another without the transfer of mass. Waves can take many forms, including electromagnetic radiation, sound waves, water waves, seismic waves, and more.
In the derivation of the wave equation, the wave is assumed to be a small disturbance from its equilibrium state. This means that the displacement of the wave from its equilibrium position is small relative to the wavelength of the wave. This assumption allows us to linearize the equation of motion and apply Newton's second law to the system.
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9. In everyday experience, why are radio waves polarized, whereas light is not?
The light we use on a daily basis is then in an unpolarized form as a result.
Why are radio waves polarized, whereas light is not?In everyday experience, radio waves are polarized, whereas light is not, due to their different sources and interactions with the environment. Radio waves are generated by antennas, which emit waves in a specific direction and with a fixed orientation.
This causes the electric field of the radio waves to have a particular alignment, resulting in polarization. On the other hand, light comes from a variety of sources, like the sun, lamps, and LED screens, which emit light in all directions and with random orientations of the electric field.
This results in an unpolarized state for the light we encounter in everyday situations.
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Compare the force-time curve for the inelastic collision to that for the nearly elastic collision.
In an inelastic collision, the objects involved stick together after impact, causing a greater deformation and a longer duration of contact. This results in a relatively wider force-time curve with a lower peak force.
When we talk about collisions, we often use force-time curves to understand the behavior of the objects involved. In an inelastic collision, the objects involved stick together after colliding, meaning that some energy is lost as heat or sound. On the other hand, in a nearly elastic collision, the objects bounce off each other and conserve most of their kinetic energy.
Comparing the force-time curves for these two types of collisions, we would expect to see some differences. In an inelastic collision, the force-time curve will generally show a larger force over a longer period of time compared to a nearly elastic collision. This is because in an inelastic collision, the objects are deforming as they stick together, creating a larger force as they do so.
In a nearly elastic collision, the force-time curve will show a shorter duration of high force, as the objects bounce off each other and transfer most of their kinetic energy without deforming. The curve will also show a more gradual decrease in force as the objects separate, since there is less energy being dissipated as heat or sound.
Overall, the force-time curves for inelastic and nearly elastic collisions will look different due to the different ways that energy is transferred and dissipated during the collision.
In a nearly elastic collision, the objects rebound after impact with minimal deformation, leading to a shorter contact duration. This produces a steeper, narrower force-time curve with a higher peak force.
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After the Sun leaves the constellation Sagittarius, how long until it returns to this constellation?
After the Sun leaves the constellation Sagittarius, it takes approximately one year, or 365.25 days, for it to return to this constellation.
This is because the Sun follows a path through the 12 zodiac constellations over the course of one year, known as the ecliptic. As the Earth orbits the Sun, the Sun appears to move through each of these constellations in turn.
Sagittarius is one of the constellations along this path, and the Sun typically passes through it between November 22 and December 21 each year. Once the Sun moves past Sagittarius, it continues on its journey through the remaining zodiac constellations until it completes its orbit and returns to Sagittarius approximately one year later.
It is worth noting that due to variations in the Earth's orbit, the exact dates on which the Sun enters and leaves each zodiac constellation can vary slightly from year to year. However, the overall period of one year remains consistent, meaning that the Sun will always return to Sagittarius approximately one year after it last passed through the constellation.
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Describe the photoelectric effect and explain why it could not be explained by Newtonian physics.
When light wave of particular frequency falls on a object this occurrence of event is photoelectric effect. This effect then stimulates emission of electron from the object.
PhenomenonPhotoelectric phenomenon cannot be defended by Newtonian physics. The energy in Newtonian physics is not continuous whereas, energy in quantum physics is continuous. Photons a energy particle which makes up the light. When collision between photon and metal surface occurs the energy of photon is absorbed by the metal electron and then this electron gets released. The debate between a wave-particle dual idea which makes up light goes on. However, light has both property of wave and particle. The photoelectric notion supports light's particle nature that is when it collides with the metal it behaves as particle. Electrons are pushed to a specific level in the line spectra where reaction between photons and electrons takes place which results in change in momentum which proves particle nature.For more information on photoelectric effect kindly visit to
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Complete question : A. Describe the photoelectric effect and explain why it could not be explained by Newtonian physics.
b. Explain what is meant by quantized light, and discuss why the photoelectric effect provides evidence for it.
c. Write 2 - 3 sentences explaining how quantum mechanics describes light and matter. How do the distinct lines in the emission spectra of elements support the idea that light can behave as a particle?
Based on the information in the passage and the data in Table 1, the observed activity in the H232R variant is most likely the result of Arg:Phosphorylated histidine is negative, and Arg would be positive
The relationship between the H232R variant, phosphorylated histidine and arginine, and protein activity.
Without additional information or context, it is not possible to determine whether the statement is true or false.
The statement appears to be referring to a scientific study that involves a variant of a protein (possibly a kinase) with a mutation at position 232 (H232R). The study appears to have investigated the activity of the variant in the presence or absence of phosphorylated histidine and/or arginine.
Table 1 likely contains data or results from the study, but it is not provided in this question, so we cannot use it to determine the veracity of the statement.
Additionally, the statement itself is incomplete and somewhat unclear. It is not clear what is meant by "the observed activity" or what specific experiment or measurement is being referred to. It is also not clear how the presence or absence of phosphorylated histidine and/or arginine relates to the activity of the H232R variant.
Without more information or context, it is not possible to determine whether the statement is true or false, or to provide a detailed explanation of the relationship between the H232R variant, phosphorylated histidine and arginine, and protein activity.
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How fast would you have to be going to completely stop the sports car? Explain the reasons for your prediction.
The initial speed, then account for factors such as the coefficient of friction and braking force. These will affect the deceleration, which in turn determines the stopping distance. The faster the initial speed, the longer the stopping distance, and vice versa
To determine how fast a sports car needs to be going to come to a complete stop, we must consider the factors that influence its stopping distance. These factors include the initial speed of the car, the coefficient of friction between the tires and the road, and the braking force applied.
Step 1: Determine initial speed (v₀) - This is the speed at which the car is traveling before braking begins.
Step 2: Calculate the deceleration (a) - The braking force (F) is determined by multiplying the mass of the car (m) by its deceleration (a). The braking force also equals the product of the coefficient of friction (µ) and the car's weight (W = m * g, where g is the acceleration due to gravity).
Thus, F = µ * W = m * a. Rearranging the formula, a = µ * g.
Step 3: Calculate stopping distance (d) - Using the equations of motion, stopping distance can be determined as
d = (v₀^2) / (2 * a).
Step 4: Analyze the prediction - With a higher initial speed, stopping distance increases. A larger coefficient of friction or stronger braking force results in a shorter stopping distance.
In summary, to completely stop a sports car, you must first determine the initial speed, then account for factors such as the coefficient of friction and braking force. These will affect the deceleration, which in turn determines the stopping distance. The faster the initial speed, the longer the stopping distance, and vice versa. Moreover, a greater coefficient of friction or stronger braking force will shorten the stopping distance.
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A gyroscope has a moment of inertia of 0.14kg×m2 and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30s. What is the value of the average frictional torque?
a.
3.3 ´ 10-2N×m
b.
8.1 ´ 10-2N×m
c.
14 ´ 10-2N×m
d.
7.0 ´ 10-2N×m
A gyroscope has a moment of inertia of 0.14kg×m2 and an initial angular speed of 15 rad/s. Friction in the bearings causes its speed to reduce to zero in 30s.The closest answer choice to this value is 3.3 * 10^-2 N*cm, so the answer is (a).
The average frictional torque formula is
α = τ / I
Where α is the angular acceleration, τ is the torque, and I is the moment of inertia.
We can rearrange this formula to solve for τ:
τ = α * I
The gyroscope starts with an initial angular speed of 15 rad/s, and comes to a stop after 30 seconds. We can use this information to find the angular acceleration:
α = (0 - 15 rad/s) / 30 s
α = -0.5 rad/s^2
Now we can use the moment of inertia given (0.14 kg*m^2) and the calculated angular acceleration to find the frictional torque:
τ = (-0.5 rad/s^2) * (0.14 kg*m^2)
τ = -0.07 N*m
The frictional torque is negative because it is acting in the opposite direction of the gyroscope's initial angular velocity. To find the average frictional torque, we can assume that the frictional torque is constant over the 30 seconds that it takes for the gyroscope to come to a stop:
τ_avg = τ / t
τ_avg = (-0.07 N*m) / (30 s)
τ_avg = -2.33 * 10^-3 N*m
We need to convert this to N*cm to match the units in the answer choices:
τ_avg = -2.33 * 10^-3 N*m * (100 cm / 1 m)
τ_avg = -0.233 N*cm
The closest answer choice to this value is 3.3 * 10^-2 N*cm, so the answer is (a).
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just by looking the above answers, when does the voltmeter shows deflection (or detectsthe light), on a sunny day or at night?
Based on the terms provided, a voltmeter will show deflection (or detect light) on a sunny day, as it receives more light exposure compared to nighttime. This deflection indicates the presence of voltage generated due to the light energy.
A voltmeter is an instrument used for measuring electric potential difference between two points in an electric circuit. It is connected in parallel. It usually has a high resistance so that it takes negligible current from the circuit. The voltmeter would show deflection (or detect the light) during a sunny day when there is sufficient sunlight to generate a voltage or electric current that can be measured by the voltmeter. At night, there would be little to no sunlight, so the voltmeter would not show any deflection or detect any light.
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A 4.0 m radius circular platform rotates with a constant angular acceleration of 20.0 rad/s^2. What is the acceleration of a point on the edge of the disk at the instant that its angular speed is 1.0 rev/s?
The acceleration of a point on the edge of a circular platform with a 4.0 m radius is 127 m/s² when its angular velocity is 1.0 rev/s and has a constant angular acceleration of 20.0 rad/s².
How to find the acceleration of a point?We can use the kinematic equations of rotational motion to solve this problem.
The first kinematic equation for rotational motion is:
ωf = ωi + αt
where ωf is the final angular velocity, ωi is the initial angular velocity, α is the angular acceleration, and t is the time.
We can use this equation to find the time it takes for the platform to reach an angular velocity of 1.0 rev/s:
1.0 rev/s = ωi + (20.0 rad/s²)t
ωi = 0 (initially at rest)
t = 1.0 rev/s / 20.0 rad/s²
t = 0.05 s
Next, we can use the second kinematic equation for rotational motion:
θ = ωit + 1/2 αt²
where θ is the angular displacement.
We can use this equation to find the angular displacement of a point on the edge of the platform during the time it takes to reach an angular velocity of 1.0 rev/s:
θ = (0)(0.05 s) + 1/2 (20.0 rad/s²)(0.05 s)²
θ = 0.0125 rad
Finally, we can use the third kinematic equation for rotational motion:
ωf² = ωi² + 2αθ
We can use this equation to find the acceleration of a point on the edge of the platform at the instant its angular velocity is 1.0 rev/s:
ωf = 1.0 rev/s = 2π rad/s
ωi = 0
θ = 0.0125 rad
a = (2π rad/s)² / 2(0.0125 rad)
a = 127 m/s²
Therefore, the acceleration of a point on the edge of the platform at the instant its angular velocity is 1.0 rev/s is 127 m/s².
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Which answer has the colors in order from the lowest frequency to the highest?
Entry field with correct answer
Red Blue Green
Blue Red Green
Blue Green Red
Red Green Blue
Red Green Blue has the colors in order from the lowest frequency to the highest. Hence option D is correct.
Visible light spectrum is nothing but the range of wavelength of radiation from 4000 angstrom to 7000 angstrom(Violet to Red). light is a energy packet. Every Photon having different wavelength travels with same velocity c (velocity of light). When we focus numbers of colors from visible spectrum to a point, that point appears as a white light. hence white light is composed of numbers of Colors in it.
Red light has wavelength 7000 angstrom and frequency 4.62 × 10¹⁴ Hz,
Green light has wavelength 550 angstrom and frequency 5.45 × 10¹⁴ Hz,
Blue light has wavelength 450 angstrom and frequency 6.66 × 10¹⁴ Hz,
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what type of isostatic forces are likely to be affecting this area at the present time? explain your answer
Isostatic forces are the result of the equilibrium between the weight of the Earth's crust and the underlying mantle. In this area, the type of isostatic forces that are likely to be affecting it at the present time depends on the local geological features.
For example, if the area has recently experienced glaciation, the weight of the ice would have caused the crust to depress, and the mantle to flow inwards to fill the space. As the ice melted, the crust would rebound upwards due to the removal of the weight, resulting in a period of isostatic uplift.
Similarly, if the area is situated on a tectonic plate boundary, the movement of the plates can cause the crust to either uplift or subside, depending on the direction and magnitude of the movement.
Other factors that can affect isostatic forces include the presence of sedimentary basins or volcanic activity.
Overall, the specific type of isostatic force affecting this area at present depends on the local geology and tectonic activity.
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A spring with a constant K is pulled back a distance X and allowed to oscillate with a period T on a horizontal frictionless surface. As the mass passes through the equilibrium positions the force of the spring is
As the mass passes through the equilibrium positions the force of the spring is increasing.
Simple harmonic motion is a specific kind of periodic motion of a body that arises from a dynamic equilibrium between an inertial force that is proportional to the body's acceleration away from the static equilibrium position and a restoring force on the moving object that is directly proportional to the magnitude of the object's displacement and acts towards the object's equilibrium position. Oscillating spring perform SHM.
as the mass passes through the equilibrium position, at the equilibrium position the force is zero and it increases with increase in displacement x according to the relation F = kx.
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For an isotropic point source of light, how does the light's intensity depend on distance from the source?
It depends on the inverse of the distance.
It depends on the inverse of the square of the distance.
It depends on the inverse of the cube of the distance.
For an isotropic point source of light, the light's intensity depend on distance from the source b. it depends on the inverse of the square of the distance
This means that as the distance from the light source increases, the intensity of the light decreases by the square of that distance. This relationship is known as the Inverse Square Law and is applicable to isotropic sources, which emit light equally in all directions. The Inverse Square Law states that the intensity (I) of the light is inversely proportional to the square of the distance (d) from the source, which can be mathematically represented as I = k / d², where k is a constant factor.
The other options mentioned, such as the inverse of the distance and the inverse of the cube of the distance, do not accurately describe the relationship between light intensity and distance for an isotropic point source. Understanding this relationship is essential in various applications, such as designing lighting systems, calculating exposure in photography, and analyzing radiation safety in the context of nuclear power plants. For an isotropic point source of light, the light's intensity depend on distance from the source b. it depends on the inverse of the square of the distance.
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What is the opposition to the establishment of magnetic lines of force in a magnetic circuit called
The opposition to the establishment of magnetic lines of force in a magnetic circuit is called magnetic reluctance. It is similar to electrical resistance in an electrical circuit, which opposes the flow of electrical current.
Magnetic reluctance is a measure of how difficult it is for magnetic flux to flow through a material or a path. The unit of magnetic reluctance is the ampere-turns per weber (AT/Wb).
The amount of magnetic reluctance depends on the material properties of the magnetic circuit, such as its dimensions, composition, and magnetic permeability. Materials with high magnetic permeability, such as iron and steel, have low magnetic reluctance and can conduct magnetic flux more easily. On the other hand, materials with low magnetic permeability, such as air or vacuum, have high magnetic reluctance and offer greater opposition to the flow of magnetic flux.
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What is the difference between convergent and parallel projection? List some different types under each category.
Convergent and parallel projections are two different ways to represent three-dimensional objects in two-dimensional space. In convergent projection, lines converge at a vanishing point, creating the illusion of depth and perspective. Parallel projection, on the other hand, maintains a consistent distance between all points in the object, resulting in a flat, two-dimensional representation.
Under the category of convergent projection, there are two main types: one-point perspective and two-point perspective. One-point perspective involves drawing an object from a single, central viewpoint, with all lines receding toward a single vanishing point on the horizon. Two-point perspective, on the other hand, uses two vanishing points to create the illusion of depth, with lines receding in different directions.
Under the category of parallel projection, there are also two main types: orthographic projection and isometric projection. Orthographic projection involves projecting an object onto a plane at a right angle to the object, resulting in a flat, two-dimensional representation that maintains accurate proportions. Isometric projection, on the other hand, uses a 30-degree angle to create the illusion of depth while still maintaining a consistent distance between all points in the object.
In summary, the main difference between convergent and parallel projection is that convergent projection creates the illusion of depth and perspective, while parallel projection maintains a consistent distance between all points in the object. There are different types of each category, including one-point and two-point perspective under convergent projection, and orthographic and isometric projection under parallel projection.
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A 0.200kg ball is shot out of a toy gun from a height of 3.00m. The gun works by compressing a spring k-175N/m a distance of 0.400m.
A. What is the force applied to the ball?
B. What is the velocity with which the ball hits the ground?
A. The force applied to the ball: 1.96 and B. The velocity with which the ball hits the ground is remains constant.
A. The force applied to the ball:
F = ma
F = 0.200 * 9.8
a= g =9.8
F = 0.200 * 9.8
F = 1.96
B. The velocity with which the ball hits the ground is remains constant. Gravity's acceleration is constantly downward and constant, however the speed and direction of the acceleration vary. The ball has zero velocity at its greatest point in its journey, and as it descends back toward the earth, its magnitude of velocity grows once more. In a uniform circular motion, velocity is constant, whereas in a non-uniform circular motion, it changes.
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The force applied to the ball is 35 N, and the velocity with which the ball hits the ground is approximately 8.4 m/s.
What is the magnitude of the applied force, and what is the final velocity of the ball?To determine the force applied to the ball and its final velocity, we can utilize the principles of conservation of energy and projectile motion.
Step 1: Calculate the force applied to the ball:
The potential energy stored in the compressed spring is converted into the kinetic energy of the ball. The potential energy of the spring can be calculated using the formula PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement of the spring.
PE = 1/2 * 175 N/m * (0.400 m)^2
PE ≈ 14 J
Since energy is conserved, this potential energy is equal to the kinetic energy of the ball:
KE = 1/2 * m * v^2
14 J = 1/2 * 0.200 kg * v^2
Solving for the velocity:
v^2 = 14 J / (0.200 kg * 1/2)
v^2 = 140 m^2/s^2
v ≈ √140
v ≈ 11.8 m/s
Step 2: Calculate the velocity with which the ball hits the ground:
Considering the vertical motion of the ball, we can use the equation of motion for free fall to determine its final velocity. The ball is initially at a height of 3.00 m, and we assume no air resistance.
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (3.00 m):
v^2 = 0 + 2 * (-9.8 m/s^2) * (-3.00 m)
v^2 = 58.8 m^2/s^2
v ≈ √58.8
v ≈ 7.7 m/s (rounded to one decimal place)
The velocity with which the ball hits the ground is the horizontal component of its velocity, which is the same as the magnitude of its velocity, approximately 7.7 m/s.
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A 65.0 kg skier slides down a 37.2 degree slope with uk = 0.107. What is the skier's acceleration and the friction force?
The skier's acceleration and the friction force are found to be 3.69 m/s² 254 N respectively.
The skier's acceleration down the slope can be determined by resolving the forces acting on the skier along the slope and perpendicular to it.
The force acting down the slope is the component of gravity, given by:
F₁ = mgsin(Ф)
= 65.09.8sin(37.2)
= 392.3 N
The force acting perpendicular to the slope is the normal force, given by,
F₂ = mgcos(Ф)
= 65.09.8cos(37.2)
= 513.9 N
The friction force is given by the product of the coefficient of kinetic friction and the normal force, i.e.,
f = uk(F₂)
Plugging in the values, we get,
f = 0.107*513.9
= 54.9 N
The net force acting down the slope is the difference between the force down the slope and the friction force:
Fnet = F₁ - f
= 392.3 - 54.9
= 337.4 N
Using Newton's second law, we can calculate the skier's acceleration down the slope,
Fnet = ma, thus,
a = Fnet/m
= 337.4/65.0
= 3.69 m/s²
Therefore, the skier's acceleration down the slope is 3.69 m/s² and the friction force is 54.9 N.
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measured independently at the position of a receiver, two speakers are measured to have sound levels of 87.0 db and 83.6 db, respectively. when both speakers emit sound simultaneously, what sound level is detected?
When both speakers emit sound simultaneously, the detected sound level is approximately 87.91 dB.
To find the combined sound level detected when both speakers emit sound simultaneously, we can use the following steps:
1. Convert the individual sound levels from decibels (dB) to their intensity ratios. Use the formula I = 10^(L/10), where L is the sound level in decibels and I is the intensity ratio.
For Speaker 1:
I1 = 10^(87.0/10) = 10^8.7 ≈ 1.995 × 10^8
For Speaker 2:
I2 = 10^(83.6/10) = 10^8.36 ≈ 4.570 × 10^7
2. Add the intensity ratios to find the combined intensity ratio:
I_total = I1 + I2 = 1.995 × 10^8 + 4.570 × 10^7 ≈ 2.452 × 10^8
3. Convert the combined intensity ratio back to decibels using the formula L = 10 * log10(I), where L is the sound level in decibels and I is the intensity ratio.
L_total = 10 * log10(2.452 × 10^8) ≈ 87.91 dB
When both speakers emit sound simultaneously, the detected sound level is approximately 87.91 dB.
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(E) The net charge on the two spheres is +Q so when they touch and separate, the charge on each sphere (divided equally) is ½ Q. F â Q1Q2 so before contact F â (2Q)(Q) = 2Q^2 and after contact F â (½ Q)(½ Q) = ¼ Q^2 or 1/8 of the original force^
Two identical conducting spheres are charged to +2Q and -Q. respectively, and are separated by a distance d (much greater than the radii of the spheres) as shown above. The magnitude of the force of attraction on the left sphere is F1. After the two spheres are made to touch and then are reseparated by distance d, the magnitude of the force on the left sphere is F2. Which of the following relationships is correct?
(A) 2Fâ = Fâ (B) Fâ = Fâ (C) Fâ = 2Fâ (D) Fâ = 4Fâ (E) Fâ = 8Fâ
The relationship is F2 = 1/8 F1. The correct option is E.
When the two spheres are separated by distance d, the electrostatic force of attraction between them is given by Coulomb's law:
F1 = (k * Q^2) / d^2
where k is the Coulomb constant.
When the spheres are touched and then separated again by distance d, they will share their charges, so each sphere will have a charge of +Q/2 and -Q/2. The electrostatic force between them will be:
F2 = (k * (+Q/2)^2) / d^2 = (k * Q^2) / (4 * d^2)
Therefore, the ratio of F2 to F1 is:
F2/F1 = [(k * Q^2) / (4 * d^2)] / [(k * Q^2) / d^2] = 1/4
So, F2 is 1/4 of F1.
However, the question asks for the relationship between F2 and F1 in terms of the force on the left sphere (not the total force of attraction between the spheres). When the spheres are separated, the force on the left sphere is F1/2, since the two spheres are identical and the force is evenly distributed between them. Similarly, when the spheres are re-separated, the force on the left sphere is F2/2. Therefore, the relationship between F2 and F1 in terms of the force on the left sphere is:
F2/2 = (1/8) (F1/2)
or
F2 = (1/8) F1
The other options are not true because:
(A) 2F1 = F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so 2F1 would be greater than F1, not equal to it.
(B) F1 = F1 - This is always true, but it does not provide any information about the relationship between F2 and F1.
(C) F1 = 2F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so F1 would be greater than 2F1, not equal to it.
(D) F1 = 4F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so F1 would be greater than 4F1, not equal to it.
Therefore, the correct option is (E) F2 = 1/8 F1.
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if two successive harmonics of a vibrating string are 360 hz and 420 hz , what is the frequency of the fundamental?
If two successive harmonics of a vibrating string are 360 hz and 420 hz.
To find the frequency of the fundamental of a vibrating string, we need to find the lowest frequency that the string can vibrate at, which is also called the first harmonic or the fundamental frequency.
We can use the formula
f = (n * v) / (2L)
Where f is the frequency, n is the harmonic number, v is the velocity of the wave, and L is the length of the string.
For the fundamental frequency, n = 1 we have
f1 = (1 * v) / (2L)
We are given two successive harmonics 360 Hz and 420 Hz. The difference between their frequencies is the frequency of one vibration cycle.
420 Hz - 360 Hz = 60 Hz
This means that the string vibrates 60 times per second between the 360 Hz and 420 Hz harmonics.
We can use this information to find the velocity of the wave, we get
v = frequency * wavelength
The wavelength is twice the length of the string for the fundamental frequency.
Wavelength = 2L
Substituting these values for v and wavelength in the above formula, we get
v = (420 Hz - 360 Hz) * (2L) = 120 Hz * (2L)
Now we can use the formula for the fundamental frequency, we get
f1 = (1 * v) / (2L)
By putting the values, we get
f1 = (1 * 120 Hz * (2L)) / (2L) = 120 Hz
Hence, the frequency of the fundamental of the vibrating string is 120 Hz.
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A thermodynamic process that happens very
quickly tends to be:
a. isovolumetric
b. adiabatic
c. isobaric
d. isothermal
The thermodynamic process happens very quickly in the Adiabatic process. Thus, option B is correct.
The thermodynamic process is the process that defines the movement of the heat flow in the system. Thermodynamics gives the relationship between heat, work, energy, and temperature. It also gives the relation between pressure, volume, and temperature.
In the isobaric process, the pressure remains constant in the system and the volume of the system changes. In the isothermal process, the temperature of the system remains constant. In the isovolumetric process, the volume remains constant, and pressure and temperature will change.
In the adiabatic process, there is no flow of heat energy between the system. This tends the thermodynamic process to proceed faster than the other.
Thus, the ideal solution is option B.
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Astronomers use the gravitational wobble of a star to indicate the presence of planets outside our solar system. Astronomers specifically look for planets in the habitable zone of 1.6x10^11 m away from the star. Calculate the mass of a planet that exerts a 1.5 x 10^13 force on a 5.4 x 10^30 kg star.
The mass of the planet that exerts a 1.5 x 10^13 N force on a 5.4 x 10^30 kg star is approximately 1.34 x 10^25 kg.
To calculate the mass of a planet, we can use Newton's law of universal gravitation. The formula is:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.674 × 10^-11 N*(m/kg)^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
In this case, F = 1.5 x 10^13 N, m1 (mass of the star) = 5.4 x 10^30 kg, and r = 1.6 x 10^11 m. We want to find m2, the mass of the planet. Rearrange the formula to solve for m2:
m2 = (F * r^2) / (G * m1)
Now plug in the given values:
m2 = (1.5 x 10^13 N * (1.6 x 10^11 m)^2) / (6.674 × 10^-11 N*(m/kg)^2 * 5.4 x 10^30 kg)
After performing the calculations, you get:
m2 ≈ 1.34 x 10^25 kg
So, the mass of the planet that exerts a 1.5 x 10^13 N force on a 5.4 x 10^30 kg star is approximately 1.34 x 10^25 kg.
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A car traveling in a straight line at 14 m/s when the driver puts on the brakes and comes to a stop in 35 m. How much time did it take for the car to come to a stop?
To calculate the time it took for the car to come to a stop, we can use the formula:
distance = (initial velocity x time) + (0.5 x acceleration x time^2)
In this case, the initial velocity (v) is 14 m/s, the distance (d) is 35 m, and the acceleration (a) is the deceleration due to braking, which is typically around -9.8 m/s^2.
We can rearrange the formula to solve for time:
time = (sqrt(2ad + v^2) - v) / a
Plugging in the values, we get:
time = (sqrt(2(-9.8)(35) + 14^2) - 14) / -9.8
time = (sqrt(-686 + 196) - 14) / -9.8
time = (sqrt(490) - 14) / -9.8
time = (22.14 - 14) / -9.8
time = 0.84 seconds
Therefore, it took the car 0.84 seconds to come to a stop.
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A simple pendulum has a mass of 0.25 kg and a length of 1.0 m. It is displaced through an angle of 30° and then released. After a time, the maximum angle of swing is only 10°. How much energy has been lost to friction?
The energy will be lost due to the resistance of the friction and hence the maximum swing angle reduces. The loss in energy will be 0.2910 J.
Given:
Mass of simple pendulum, m= 0.25 kg.
Length of the simple pendulum, l = 1.0 m.
The potential energy for the given two cases will be different, hence we can calculate the loss in energy by applying the law of conservation of energy which states that the total energy for a system remains the same.
For the first case, angular displacement θ₁ = 30⁰
The height of the pendulum from the mean position is given by
h₁ = l×(1-cos30⁰)
Energy, E₁ = mgh₁
E₁= 0.25 × 9.8 × 1.0 × (1 - cos30⁰)
For the second case, angular displacement θ₂ = 10⁰
The height of the pendulum from the mean position is given by
h₂ = l×(1-cos10⁰)
Energy, E₂ = mgh₂
E₂ = 0.25 × 9.8 × 1.0 × (1 - cos10⁰)
From the law of conservation of energy
Initial energy = final energy + losses
Hence,
E₁ = E₂ + ΔE
ΔE = E₁ - E₂
ΔE = mgh₁ - mgh₂ = mgl(cos10⁰ - cos30⁰)
ΔE = 0.25 × 9.8 × 1.0 × (cos10⁰ - cos30⁰)
ΔE = 0.2910 J
Therefore, the energy lost due to friction is 0.2910 J.
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the amount of time it takes an object to reach its max height is the same time it takes for the object to ___________.
The amount of time it takes an object to reach its max height is the same time it takes for the object to fall back down to its original height.
This is the time it takes for the object to travel from its launch point to its peak height and then back down to its initial launch point. During this time, the object is under the influence of gravity, so its velocity and acceleration change. At the start, the object accelerates towards its maximum height, then its velocity decreases as it approaches its maximum height, and finally the object decelerates as it returns to its launch point.
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STT 10 A car pulls away from a stop sign with a constant acceleration. After traveling 10 m, its speed is 5 m/s. What will its speed be after traveling 40 m?A 10 m/s B 20 m/sC 30 m/s D 40 m/s
The speed of the car after traveling 40 m is 30 m/s. So the correct answer is : C.
We can use the following kinematic equation to solve this problem:
[tex]v^2 = u^2 + 2as[/tex]
We know that the car starts from rest, so u = 0. Also, the acceleration is constant throughout the motion. We are given that the car has a constant acceleration and after traveling 10 m, its speed is 5 m/s. Using these values, we can solve for the acceleration:
[tex]v^2 = u^2 + 2as \\5^2 = 0^2 + 2a(10) \\a = 12.5 m/s^2[/tex]
Now we can use the same equation to find the final velocity of the car after traveling 40 m:
[tex]v^2 = u^2 + 2as \\v^2 = 0^2 + 2(12.5)(40) \\v = 30 m/s[/tex]
Therefore option: C is correct.
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The fraction of energy carried by the reflected sound is small if the surface is
The fraction of energy carried by the reflected sound is small if the surface is smooth and hard.
This is because when sound waves hit a smooth and hard surface, they bounce back in a very predictable manner, following the law of reflection.
This means that the angle of incidence is equal to the angle of reflection, and the sound waves are directed away from the surface in a way that does not scatter or disperse the energy.
As a result, the amount of energy that is reflected is relatively small compared to the energy that is absorbed or transmitted through the surface.
In contrast, rough and soft surfaces tend to scatter and absorb sound waves, which can result in a higher fraction of energy being reflected back.
This is why soundproofing materials are often designed to be soft and porous, in order to absorb and dampen sound waves rather than reflect them back into space.
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Provide a summary of your predictions. What are the circumstances under which you predict that one object will exert a greater force on the other object?
According to Newton's laws of motion, the amount of force that one object exerts on another is directly proportional to its mass and acceleration. In other words, the greater the mass and acceleration of an object, the greater the force it will exert on another object.
Additionally, the force between two objects also depends on their distance from each other. This is described by the inverse square law, which states that the force between two objects decreases as the square of the distance between them increases. So if two objects are moved farther apart, the force between them will decrease.
It's also important to note that the force between two objects is always equal and opposite. This is known as Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Therefore, to predict which object will exert a greater force on another, we need to consider factors such as the mass and acceleration of each object, their distance from each other, and the nature of the interaction between them. These factors will determine the overall force exerted and the direction of that force.
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suppose a rocket in outer space is thrust along the y direction with an acceleration of 15 m/s^2 while drifting freely (no applied force) in the x direction. What is the path followed by the rocket?
The path followed by the rocket is a circular path.
Since, the rocket is thrust along the y-direction, we can say that its velocity is in the y-direction.
The direction of acceleration is along the x-direction.
Therefore, it can be said that the velocity and acceleration of the rocket are in a mutually perpendicular direction. Also, given that no external force is acting.
Therefore, the rocket is in circular motion.
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