The following is the order of events that take place when oceanic plates spread apart at the mid-ocean ridge:
1. Basaltic lava erupts from the rift.
2. Blocks of oceanic crust are down-dropped in normal faults.
3. Sediment settles onto new basalt.
4. Oceanic crust has a smooth surface covered by layered sediment.
When an oceanic plate diverges from another oceanic plate or diverges from a continental plate, a mid-ocean ridge is created.
The mid-ocean ridge system is formed as magma rises from the mantle and is injected into the crustal rocks above.
The injection of magma creates a rift and extrusion of basaltic lava along the crest of the ridge.
The lava flows out from the crest of the ridge and cools and solidifies to form new crustal rock.
Thus, the first event that occurs when oceanic plates spread apart at the mid-ocean ridge is that basaltic lava erupts from the rift, and blocks of oceanic crust are down-dropped in normal faults.
As the new basaltic rock cools and solidifies, the sediment settles onto it. Finally, the oceanic crust has a smooth surface covered by layered sediment.
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A square loop of wire is carrying current in the counterclockwise direction. There is a horizontal uniform magnetic field pointing to the right.Question 1: What is the direction of the net force on the loop?(A) out of the screen(B) into the screen(C) the net force on the loop is zeroQuestion
If the magnetic field and the velocity are perpendicular, the force is maximum, and if they are parallel, the force is zero. The direction of the magnetic force can be determined using Fleming’s left-hand rule. The thumb represents the direction of the motion of the charge, the first finger represents the direction of the magnetic field, and the middle finger represents the direction of the magnetic force.
A square loop of wire carrying current in the counterclockwise direction will experience a force.
The force will be in the direction given by Fleming’s left-hand rule. The magnetic field is uniform and horizontal, and it is pointing towards the right. The question is asking for the direction of the net force on the loop. The direction of the net force on the loop can be determined using the right-hand palm rule.
The right-hand palm rule states that the thumb represents the direction of the current, and the fingers represent the direction of the magnetic field. If the fingers of the right hand are curled in the direction of the magnetic field and the thumb in the direction of the current, then the direction of the force is given by the palm.
In this case, the palm points upwards, which means that the net force on the loop is out of the screen. Therefore, the correct option is (A) out of the screen. Magnetic force The force exerted on a charged particle moving in a magnetic field is known as magnetic force. The direction of the magnetic force on the moving charge is perpendicular to the plane formed by the magnetic field and the velocity of the charge.
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Describe such a motion from every day experience of deceleration to acceleration
Answer:
Throwing a ball up into the air. The ball will going accelerate up, then slowing down due to gravity, briefly stop, and then accelerating on its way down to the floor.
Explanation:
which of the following is true of polarizable electrodes? group of answer choices current passes freely across the electrode-electrolyte interface, requiring no energy to make the transition. polarizable electrodes are used for stimulation. no actual charge crosses the electrode-electrolyte interface when a current is applied. a and b
The true statement about polarizable electrodes is: No actual charge crosses the electrode-electrolyte interface when a current is applied. This is the correct option among the given options.
In the case of polarizable electrodes, no actual charge crosses the electrode-electrolyte interface when a current is applied.
What is the meaning of polarizable electrodes?Polarizable electrodes are those electrodes which are chemically reversible, so they can store electrical energy as well as release it.
An electrode is a metal strip that conducts electricity into or out of a solution. Polarization happens at the interface of the electrode and the electrolyte solution. The potential difference created in the electrode-electrolyte interface causes this phenomenon.
How is the electrode polarization related to the efficiency of a battery?The efficiency of a battery is inversely proportional to electrode polarization. Polarization happens due to the formation of reaction intermediates on the electrode surface, which lowers the reaction rate. The amount of polarization also depends on the electrode surface's area and the current flow. Because of this, polarization causes a reduction in current efficiency.
Polarizable electrodes are used in stimulation in a variety of ways. Polarization, on the other hand, occurs when an electrode is used for prolonged periods. The electrode becomes inert over time, and it loses its ability to conduct a charge because of polarization. As a result, the life of the electrode is shortened.
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Articulate succinctly (and use a sketch if helpful) why the sign of the wave-function matters when two or more atoms form bonds.
The sign of the wave-function is crucial to understand the chemical bonding of two or more atoms. It is responsible for providing stability and the formation of covalent bonds. Covalent bonds are characterized by the sharing of electrons between two or more atoms to achieve a stable state.
What is Covalent bonds?The two hydrogen atoms share an electron in their 1s orbital. When the two atoms approach each other, their 1s orbitals overlap, and the wave-function of each electron combines. This combination of wave-function occurs because of the Schrödinger wave equation.
The significance of the sign of the wave-function is that it determines the probability of an electron's presence in a particular area around the nucleus of an atom. The Schrödinger wave equation is sensitive to the sign of the wave-function because the wave-function squared gives the probability density of the electron's presence.
Therefore, when two atoms come together to form a bond, the sign of the wave-function becomes critical. If the signs of the wave-function for the two hydrogen atoms are the same, the probability of the two electrons sharing space increases, which results in a stable molecule. If the signs of the wave-function are different, the probability of electron sharing decreases, which results in an unstable molecule.
A sketch can be helpful to understand the concept of covalent bonding. When two hydrogen atoms come together to form a molecule, they share an electron in their 1s orbitals, resulting in a stable molecule. A sketch will provide a visual representation of the sharing of electrons between the two atoms.
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two springs are connected in series so that spring scale a hangs from a hook on the ceiling and a second spring scale, b, hangs from the hook at the bottom of scale a. apples weighing 287 n hang from the hook at the bottom of scale b. ignore the weights of the ropes and scales. what is the reading on the lower scale b?
The reading on the scale B is 287N. This is because the two springs in series are both in equilibrium, meaning that the forces exerted by each spring are equal to each other and to the weight of the apples (287N).
What is the reading on scale?
To determine the reading on the lower scale b, you need to calculate the total elongation of both the springs. Let us assume that the elongations of springs A and B are dA and dB, respectively. Spring Scale A is fixed to the ceiling and is vertically above Spring Scale B. Spring Scale A reads the total weight of the combination, which is the weight of both the apples and the scales. So, the weight on Scale A is 287N (Weight of the apple).
The force exerted by Scale A is divided between the two springs, so you need to know the spring constant for both the springs to calculate how the weight will be divided. Let’s assume that the spring constant for spring A is KA and spring constant for spring B is KB. Hence, we know the following:
F = kx
where, F is the force exerted by the spring, x is the elongation of the spring, and k is the spring constant.
We can express this as:
F = m×g
where, m is the mass attached to the spring and g is acceleration due to gravity.
Using the above two equations, we can get the following:
x = m×g/k
The weight on Scale B is 287N, which is the force exerted by spring B.
So, 287 = KB×dB
Also, the force exerted by Scale A is divided between the two springs. The force on spring A is the total weight, which is 287N plus the weight of the two spring scales (which can be ignored). So, the force on spring A is 287N.
So, 287 = KA×dA + KB×dB
Since both the springs are connected in series, the total elongation (d) is the sum of the elongations of the individual springs. Hence,d = dA + dB. So, substituting the value of dB in the above equation:
287 = KA× dA + KB×dA/KB
Therefore, dA = 287/ (KA + KB)
Therefore, the reading on scale B (lower scale) is: dB = 287/KB. So, the reading on scale B is 287/KB.
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a 150 kg cart on a flat surface is pulled by a force of 120 n at the 50 degrees with respect to the horizontal surface for a distance of 15 meters, what is the work done to the cart by the pulling force? ignore the friction between the cart and flat surface.
The work done to the cart by the pulling force is 16950 J.
Work is the transfer of energy that happens when a force makes an object move. To calculate the work done to an object, we multiply the force applied by the distance moved in the direction of the force
Given a force of 120 N applied at an angle of 50° with respect to the horizontal surface.
We can resolve the force into its horizontal and vertical components as follows:
Horizontal force, Fx = F cos θ = 120 cos 50° = 91.76 N
Vertical force, Fy = F sin θ = 120 sin 50° = 91.67 N
Ignoring friction, the net force acting on the cart is the horizontal force, Fx.
The acceleration produced by the force is given by:
F = ma => a = F / m => a = 91.76 / 150 = 0.611 m/s²
The displacement of the cart in the direction of the force is the same as the distance covered, which is given as 15 meters.
Therefore, the work done to the cart by the pulling force is given by: W = Fd cos θW = 91.76 × 15 × cos 50°W = 16950 J.
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A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at P and increases linearly with distance past P, reaching a value of 0.600 at 12.5 m past point P.1)Use the work-energy theorem to find how far this box slides before stopping.2)What is the coefficient of friction at the stopping point?3)How far would the box have slid if the friction coefficient didn't increase, but instead had the constant value of 0.100?
1) The box slides 23.6 m before stopping.
2) The coefficient of friction is 0.600.
3) If the friction coefficient had a constant value of 0.100, then the box would have slid a distance of 12.5 m
1)Using the work-energy theorem, we can find how far the box slides before it stops. The theorem states that the work done by all forces acting on the object is equal to the change in its kinetic energy, or:
W = ΔKE
Where W is the work done by all forces, and ΔKE is the change in kinetic energy. Since the work done by all forces is equal to the friction force, the work done by friction is equal to the change in kinetic energy. Therefore, the equation can be rewritten as:
Wfriction = ΔKE
In this situation, the friction force increases linearly with distance, so the work done by friction (Wfriction) is the integral of the friction force over the distance. The integral is equal to the area under the graph of friction force versus distance. Therefore, the equation can be rewritten as:
∫Ffriction(x)dx = ΔKE
The integral is equal to the area under the graph of friction force versus distance from 0 to the stopping point. Since the coefficient of friction increases linearly from 0.100 at P to 0.600 at 12.5 m past point P, we can calculate the stopping point using the equation:
0.100x + 0.500(x-12.5) = ΔKE
Solving the equation for x, we find that the box stops at x = 23.6 m.
2)At the stopping point, the coefficient of friction is 0.600.
3) This is because the integral of the friction force with a constant coefficient of 0.100 is equal to 0.100x, where x is the distance traveled.
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A block of mass m is at rest at the origin at t=0. It is pushed with constant force F0 from x=0 to x=Lacross a horizontal surface whose coefficient of kinetic friction is μk=μ0(1−x/L). That is, the coefficient of friction decreases from μ0 at x=0 to zero at x=L.
Part A
We would like to know the velocity of the block when it reaches some position x. Finding this requires an integration. However, acceleration is defined as a derivative with respect to time, which leads to integrals with respect to time, but the force is given as a function of position. To get around this, use the chain rule to find an alternative definition for the acceleration ax that can be written in terms of vx and dvxdx. This is a purely mathematical exercise; it has nothing to do with the forces given in the problem statement.
Express your answer in terms of the variables vx and dvxdx.
I got the answer:
ax =
dvxdxvx
And this was correct, but Im having trouble with Part B:
Now use the result of Part A to find an expression for the block's velocity when it reaches position x=L.
Express your answer in terms of the variables L, F0, m, μ0, and appropriate constants.
To start, let's examine the forces that the block is subjected to as it moves from x=0 to x=L.
The block is at rest at the beginning of the motion (x=0), thus there is no net force acting on it. F0 is the force pushing the block, and f = k N = k mg, where N is the normal force and g is the acceleration brought on by gravity, is the force of kinetic friction acting in the opposite direction. The block is stationary, thus we have:
F0 - μ0 mg = 0
The force pushing the block must thus be equal to and in opposition to the force of friction.
The coefficient of kinetic friction changes as the block travels over the surface.
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What are the 4 factors of material resistance?
The four factors of material resistance are temperature, strain rate, stress state, and environment.
Temperature affects the flow of material, strain rate refers to the speed at which it is deformed, stress state is the amount of force applied, and environment relates to the presence of contaminants or corrosive agents.
The four factors of material resistance are temperature, time, applied stress, and strain rate. These factors are important in determining the strength and durability of a material and its ability to resist deformation or failure.Temperature: The temperature of a material can have a significant impact on its strength and resistance to deformation. Higher temperatures can cause a material to soften and weaken, while lower temperatures can make it more brittle and prone to cracking.Time: The duration of an applied load or stress can affect a material's strength and ability to resist deformation. Over time, a material may experience creep, which is a gradual deformation under a sustained load.Applied Stress: The magnitude of an applied stress or load can also affect a material's resistance to deformation. Higher stress levels can cause a material to reach its yield strength or fracture point more quickly.Strain Rate: The rate at which a material is deformed can also impact its strength and resistance to deformation. Higher strain rates can cause a material to behave differently than it would under static loading conditions, and can lead to failure at lower stress levels.
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A girl cycles a distance of 50 meters using a total force on the pedals of 150 N. Calculate the work done on the bicycle. (don't forget the units on your answer)
Answer:
7500Joules
Explanation:
workdone= force × Distance
a positively charged insulated rod is brought near two neutral conducting spheres, a and b, which are touching each other and held in place and insulated from the rest of the environment. once the rod is close to (but not touching) sphere a, the spheres are separated from each other. sphere a is then suspended from a string, and the rod is brought near it again while sphere b is moved far away. sphere a is attracted to the rod. the investigation is repeated with a negatively charged rod, and the observed results are the same. which of the following best explains why the results are the same for a positively charged rod and a negatively charged rod?
The best explanation for why the results are the same for a positively charged rod and a negatively charged rod is that the charge on the spheres is redistributed to create opposite charges on the spheres.
The charge on the spheres is redistributed to create opposite charges on the spheres, which is why the results are the same for a positively charged rod and a negatively charged rod. This redistribution happens as a result of induction. As a result of the charge redistribution, the spheres develop an attraction to the rod. When a negatively charged rod is brought close to the spheres, the charge on the spheres is redistributed, causing one of the spheres to have a net positive charge and the other to have a net negative charge.
The sphere with the opposite charge (in this case, the one with a net positive charge) is attracted to the negatively charged rod, while the sphere with the same charge (in this case, the one with a net negative charge) is repelled. This redistribution results in the spheres separating from one other.When a positively charged rod is brought near the spheres, the same charge redistribution occurs, resulting in the same attraction between the oppositely charged sphere and the rod. Sphere B is far away, hence it does not undergo any charge redistribution as a result of the presence of the charged rod.
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a bullet of mass, m is fired horizontally into a block of mass, m as shown. the block with the embedded bullet rises to height, h. acceleration due to gravity is g acting downward. part a: what is the speed, v of the block (with the bullet embedded in it) immediately after the collision, in terms of the variables provided in the problem?
The speed of the block (with the bullet embedded in it) immediately after the collision, in terms of the variables provided in the problem, is given by [tex]v = (m/(m + M)) * (2gh)^{0.5}[/tex], where m is the mass of the bullet, M is the mass of the block, and h is the height to which the block rises.
First, we assume that the collision is perfectly inelastic, meaning that the bullet becomes embedded in the block and they move together as a single mass. In this case, the conservation of momentum equation can be written as:
[tex]m_{bullet} * v_{bullet} = (m_{block} + m_{bullet}) * v_{final}[/tex]
where v_bullet is the initial velocity of the bullet, v_final is the final velocity of the block with the embedded bullet, and we have used the fact that the block and bullet move together as a single mass after the collision.
Next, we can apply conservation of energy to find the velocity of the block with the embedded bullet at the height h. Since the collision is inelastic, some of the initial kinetic energy is lost as heat and deformation. We can express the conservation of energy equation as:
[tex](1/2) * m_{bullet} * v_{bulle}t^2 = (m_{block} + m_{bullet}) * g * h[/tex]
where g is the acceleration due to gravity and we have used the fact that the potential energy gained by the block-bullet system is equal to the initial kinetic energy of the bullet.
Solving for v_final in the momentum equation and substituting it into the energy equation, we get:
[tex](1/2) * m_{bullet} * v_{bullet}^{2} = (m_{block} + m_{bullet}) * g * h[/tex]
[tex]v_{final} = v_{bullet} * (m_{bullet} / (m_{block} + m_{bullet}))^{0.5}[/tex]
So the speed of the block with the bullet embedded in it immediately after the collision can be calculated using this equation, where we plug in the values of [tex]m_{bullet}, m_{block}, v_{bullet}[/tex], and h.
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during a one-second period, air is added into a rigid tank. the volume of the tank is 3 m3 and the initial density of air is 1.2 kg/m3; at the end of the charging process, the density of air reaches 6.3 kg/m3. what is the mass flow rate of air that is entering the tank?
The mass flow rate of air that is entering the tank is 15.3 kg/s.
The mass flow rate of air that is entering the tank can be calculated by using the following formula:
Mass flow rate = density × volume flow rate
The term "density" refers to the amount of mass per unit volume. It is calculated as the mass of an object divided by its volume. Mass flow rate is the mass of a fluid that flows through a given area per unit of time.
The volume of the tank is 3 m³.
The initial density of air is 1.2 kg/m³.
At the end of the charging process, the density of air reaches 6.3 kg/m³.
We will first find the volume flow rate.
The volume flow rate is equal to the change in volume over time.
Volume flow rate = Volume change / Time taken = 3 m³ / 1 sec = 3 m³/s
Now, we can calculate the mass flow rate using the formula:
Mass flow rate = density × volume flow rate
Density = 6.3 kg/m³ − 1.2 kg/m³ = 5.1 kg/m³
Mass flow rate = 5.1 kg/m³ × 3 m³/s = 15.3 kg/s
Therefore, the mass flow rate of air entering the tank is 15.3 kg/s.
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Help would be greatly appreciated!
Answer:
V = X i + Y j expression of vector in terms of unit vectors i andj
V = 12.0 i + 9.00 j
V = (12.0^2 + 9.00^2)^1/2 = 15.0 magnitude of resultant vector
Note this is a multiple of a 3, 4, 5 right triangle
5 is the magnitude of a 3, 4, 5 right triangle the given vector is 3X the 3, 4, 5 triangle or 15
how much work is it to push a box (mass 130 kg) up an incline (angle 20 degrees with the horizontal) that is 10 meters long, if the coefficient of kinetic friction between the box and the incline is 0.5?
The amount of work required to push the box (mass 130 kg) up an incline (angle 20 degrees with the horizontal) that is 10 meters long, with the coefficient of kinetic friction being 0.5, is 25145.1 J.
The frictional force that acts on a body when it slides on a surface is referred to as kinetic friction or sliding friction. When a body slides across a surface, the force that opposes its motion is referred to as kinetic friction.Kinetic friction equationThe formula for kinetic friction can be written as follows:
F[tex]_k[/tex] = μk N
Where:
F[tex]_k[/tex] is the force of kinetic friction
N is the normal force
μk is the coefficient of kinetic friction
The force of kinetic friction is equivalent to the product of the normal force and the coefficient of kinetic friction. The normal force is equal to the weight of the object in this case, i.e., N = mg.
The force required to push the box up the incline can be calculated using the following formula:
W = mg(sin θ + μk cos θ)
Distance traveled by the box is 10 meters, so the work done to push the box up the incline is equal to
W = F.d
where,
F = force applied
d = distance moved by the box
W = 130 * 9.8 * (sin 20 + 0.5 cos 20) * 10
W = 25145.1 J
Therefore, it would take 25145.1 J of work to push a box.
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Select all of the following that would definitely increase the pressure of an ideal gas. Decreasing the volume without changing the temperature Increasing the temperature without changing the volume
Explanation:
Ideal Gas Law
PV = n R T
P = n R T / V Without changing more than one variable:
increasing n (number of moles)
Increasing T
decreasing V
Will all increase the pressure , P
Hominin remains have been found at sites throughout Africa, Europe, and Asia. Below are five of these fossil sites and their main finds. Place them in order from the oldest (earliest in fossil record = most ka) to the youngest (most recent in fossil record = least ka).
1. Omo Kibish: incomplete fossil skull (oldest)
2. Herto: a skull intermediate between H. heidelbergensis and modern H. sapiens
3. Klasies River: fragmentary hominin remains
4. Tianyuan Cave: modern human maniple and femur (youngest)
Hominin remains have been found at sites throughout Africa, Europe, and Asia. Below are five of these fossil sites and their main finds. The correct order of the given hominin fossil sites from oldest to youngest is Omo kibish, Klasies river, Herto, and Tianyuan cave.
Omo kibish is incomplete fossil skull (oldest)The oldest fossil site on the list is Omo kibish in Ethiopia, which is dated to around 195,000 years ago. The site has a partial skull, lower jawbone, and a few other fragments of the skull. Klasies river is fragmentary hominin remains Klasies river Mouth in South Africa is dated back to around 120,000 years ago. The site contains human fossils along with the remains of other animals.
Herto is a skull intermediate between H. heidelbergensis and modern H. sapiens, Herto Bouri in Ethiopia, dated to around 160,000 years ago. The site contains 3 complete hominid skulls which were much more modern than expected for their age. The skulls were similar to Homo sapiens, but with some differences. Tianyuan cave is modern human maniple and femur (youngest). Tianyuan cave in China, dated to around 40,000 years ago, contains one of the earliest modern human fossils. A complete set of human teeth and bones from a foot, leg, and hand were found at this site.
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Can anyone help me please ..I need it hurry within 6 hrs.please.
Brainliest for the first best answer.
The value of the current in the given scenario include the following:
I.) 0.67 A
0.67 Aii.) 0.25A
0.67 Aii.) 0.25Aiii.) 0 A
How to calculate the current through a circuit when the switch is either open or closed?The formula that can be used to calculate the current through a circuit I = V/R
Where V = Voltage
Voltage R= Resistance
Voltage R= Resistance I = Current
Nit's that whenever a switch is closed, current moves through the circuit but when it's open there is not net movement of current.
When switch K1 is closed :
current = V/R = 2/3 = 0.67 A
When switches K1 and K2 are both closed
current = 2/5+3 = 2/8 = 0.25A
When switch K1 is open and K2 is closed, there will be no net current are current should flow from K1 which is open.
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. determine the change in rotational kinetic energy when the rotational velocity of the turntable of a stereo system increases from 0 to 33 rpm. its rotational inertia is 6.0 * 10-3 kg
Answer: The change in rotational kinetic energy = 0.0358 J
Explanation:
The kinetic energy of a rotating rigid body is directly proportional to the moment of inertia and the square of the angular velocity, K = 1/2 I w²
According to the given information :
The rotational kinetic energy is given by E= 1/2 I w²
E1= 1/2 ×6× 10-³ ×0 = 0
w2 = 33rpm = 33 (2π/60 )= 3.456 rad/s
E2 = 1/2 ×6×10 -³ × (3.456) ² = 0.0358 J
E = E1- E2 = 0- 0.0358 J = 0.0358 J
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The high temperature microwave spectrum of KCl vapor shows an absorption at a frequency of 15376 MHz. a) Show that this frequency represents a photon with energy of 10.19 x 10-24]. b) This absorption peak has been identified with the ] = 1 7 J = 2 transition of 39K35CL. Given that the atomic masses of 39K and 3Cl are 38.96 and 34.97 g/mole, respectively, calculate the internuclear distance (i.e: bond length) of 39KBCl in meters and A Compare your answer above to the experimentally-determined bond length of HCL, 1.275A. Using concepts from general chemistry, explain why the comparison does or does not make sense_
In the comparison between these compounds, the bond is weaker due to the difference in electronegativity between the two atoms.
What is the wavelength?We can use the relation: E = hν
where, E is the energy of the photon, h is Planck's constant, and ν is the frequency of the photon.
Converting to Hz:
ν = 15376 MHz = 15376 × 106 Hz = 1.5376 × 10¹⁰ Hz
Substituting the frequency into the formula for photon energy: E = hν
E = 6.626 × 10⁻³⁴ J s × 1.5376 × 10¹⁰ Hz
E = 1.019 × 10⁻²³Joules
The frequency of the photon can be used to calculate the wavenumber, which in turn can be used to determine the internuclear distance of the molecule. The wavenumber (ν¯) of the photon is defined as the frequency divided by the speed of light, c:
ν¯= ν/c
where, c is 2.998 × 10⁸ m/s.
Converting the frequency into wavenumber:
ν¯= ν/c = 1.5376 × 10¹⁰ Hz/2.998 × 10⁸ m/s = 51.31 cm⁻¹
The wavenumber of the photon can be used to calculate the internuclear distance (r) by using the equation:
r = [h/(8π²cμ)]½ × (1/ν¯)
where, h is Planck's constant, c is the speed of light, and μ is the reduced mass of the molecule (m₁m₂/m₁ +m₂).
For K₃₅Cl₃₉, the atomic masses of K and Cl are 39 and 35, respectively. Therefore, m₁ = 39, u = 39 × 1.66 × 10⁻²⁷ kg = 6.474 × 10⁻²⁶ kg, m₂ = 35 u = 35 × 1.66 × 10⁻²⁷ kg = 5.81 × 10⁻²⁶ kg,
μ = (m₁m₂/m₁ +m₂) = 39 × 35/(39 + 35)
u = 16.86
u= 16.86 × 1.66 × 10⁻²⁷kg = 2.798 × 10⁻²⁶kg
Substituting the values of the constants and the wavenumber: r = [h/(8π²cμ)]½ × (1/ν¯)r = [(6.626 × 10⁻³⁴ J s)/(8π² × 2.998 × 10⁸ m/s × 2.798 × 10⁻²⁶ kg)]½ × (1/51.31 cm⁻¹)r = 1.873 × 10⁻¹⁰ m = 1.873 Å
We can compare this bond length to that of HCl, which is 1.275 Å. The internuclear distance of K₃₅Cl₃₉ is much longer than that of HCl, indicating that the bond in K₃₅Cl₃₉ is weaker. This is consistent with the fact that K₃₅Cl₃₉ is a heteronuclear diatomic molecule, whereas HCl is a homonuclear diatomic molecule. In a heteronuclear diatomic molecule, the bond is weaker due to the difference in electronegativity between the two atoms.
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What energy occurs when nuclear bonds split or fuse together?
Atoms are divided via nuclear fission, which is employed in power plants to liberate energy. Atoms are combined during fusion, which happens in stars like the sun, and produces energy. A source of clean energy with less radioactive waste is fusion.
Nuclear links can break or fuse together to release energy. A substantial quantity of energy is released when an atom's nucleus splits into two or smaller nuclei during nuclear fission. Electricity is produced using this method at nuclear power plants. Contrarily, nuclear fusion is the process in which two or more atomic nuclei come together to produce a heavier nucleus, releasing a massive amount of energy in the process. This happens in stars like the sun naturally when hydrogen is fused with helium to create energy. Nuclear fusion is being studied by scientists as a possible clean energy source since it generates a lot less radioactive waste than nuclear fission.
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imagine that earth was upright with no tilt. how would this affect the seasons?
Answer:
There would be no "seasons"
The person would always be subjected to the same amount of sunlight.
Since the earth is tilted at about 23 deg, a person at that latitude would be subjected to sunlight from overhead to sunlight that appears 46 deg N of vertical.
Make a drawing of the problem situation including labeled vectors to represent the motion of the block as well as the forces on it. What measurements can you make with a meter stick to determine the angle of incline?
Draw a free-body diagram of the block as it slides down the track. Choose a coordinate system that will make calculations of energy transfer to and from the block easiest. What is your reason for choosing that coordinate system?
Transfer the force vectors to your coordinate system. What angles between your force vectors and your coordinate axes are the same as the angle between the track and the table?
In the coordinate system you have chosen, is there a component of the block’s motion that can be considered as in equilibrium? Use Newton’s second law in that direction to get an equation for the normal force in terms of quantities you know or can measure. Does the normal force increase, decrease, or stay the same as the ramp angle increases?
Write down the expression for the acceleration of the block in terms of quantities you know or can measure. Is the acceleration positive, negative, or zero? What is would be the difference between a positive and negative acceleration in this case?
Sketch a graph of the frictional force as a function of the normal force if the approximate relationship between them is good in this situation. How would you determine the coefficient of kinetic friction from this graph?
Restate the problem in terms of quantities you know or can measure. Beginning with basic physics principles, show how you get equations that give the forces you need to solve the problem. Make sure that you state any approximations or assumptions that you are making. Make sure that in each case the force is given in terms of quantities you know or can measure. Write down the approximate expression for friction that you are testing and sketch a graph of frictional force as a function of normal force for that equation.
First draw a free-body diagram of the block as it slides down the track. Then choose a coordinate system that will make calculations of energy transfer to and from the block easiest, and transfer the force vectors to your chosen coordinate system.
To determine the angle of incline, you can use a meter stick to measure the length of the ramp and the vertical height of the ramp, and then use basic trigonometry to calculate the angle. The expression for the acceleration of the block in terms of quantities you know or can measure is given by:
a=Fnet/m,
where Fnet is the sum of all forces acting on the block and m is its mass. The acceleration will be positive if the net force on the block is greater than zero, negative if the net force is less than zero, and zero if the net force is zero.
Lastly, restate the problem in terms of quantities you know or can measure and use basic physics principles to show how you get equations that give the forces needed to solve the problem. Make sure that you state any approximations or assumptions that you are making, and write down the approximate expression for friction and sketch a graph of frictional force as a function of normal force for that equation.
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A beam is supported at its middle point (fulcrum). On the left of fulcrum is a box of20 kgat2.0 maway from center. On the right side, another box of40 kgis placed at1.0 mfrom the fulcrum. The beam is balanced and horizontal. What is the vertically upward normal reaction force(Fn)on the beam at the fulcrum? useg=10 m/s∧2.
260 N
375 N
560 N
600 N
The vertically upward normal reaction force on the beam at the fulcrum is 600 N. This can be calculated by taking the total moment of box. Thus, the correct option is D.
What is the vertically upward normal reaction force?The vertically upward normal reaction force on the beam at the fulcrum is 375 N. Let the normal reaction force exerted on the beam be N1, and the normal reaction force exerted by the 20 kg box be N2. Since the beam is balanced and horizontal, there must be no net force in any direction, and the sum of the moments must be zero.
Therefore, taking moments about the fulcrum, we get:
20 × 2.0 × 10 + 40 × 1.0 × 10 = N1 × 0
Hence, N1 = (20 × 2.0 × 10 + 40 × 1.0 × 10)/0 = 1200/0, which is undefined or infinity.
We can see that our equation was wrong. What we have to do is that we need to balance the moments of the two boxes by adding their moments together. The moment of the 20 kg box is:
20 × 2.0 × 10 = 400 Nm.
The moment of the 40 kg box is: 40 × 1.0 × 10 = 400 Nm as well. So, the total moment is: 400 + 400 = 800 Nm. To balance the moments, we need the fulcrum to exert an equal and opposite moment.
So, N1 × 0 = 800 Nm, which gives N1 = 0.The normal force exerted on the beam by the fulcrum is zero. Therefore, the total upward normal reaction force acting on the beam is equal to the weight of the two boxes. Thus,
Fn = (20 + 40) × 10
Fn = 600 N
Therefore, the vertically upward normal reaction force on the beam at the fulcrum is 600 N. Hence, the correct option is D.
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In the absence of external forces, momentum is conserved ina. quadrupledb. Yes, force will be less on a carpetc. The component of the weight of the block of ice that is parallel to the slope.d. in both elastic and inelastic collisions
In both elastic and inelastic collisions, momentum is conserved in the absence of external forces. Option d is the correct answer.
In the absence of external forces (such as friction, air resistance, or other external influences), the total momentum of a system remains constant. This is known as the law of conservation of momentum. It applies to all types of collisions, including elastic and inelastic collisions.
In an elastic collision, the total kinetic energy of the system is conserved, in addition to the momentum. In an inelastic collision, some of the kinetic energy is transformed into other forms of energy (such as heat or deformation), but the total momentum is still conserved. The conservation of momentum is a fundamental principle in physics and has many applications, from understanding the behavior of subatomic particles to predicting the trajectories of spacecraft. Hence option d is correct.
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A two-stage rocket is traveling at 1210m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 40m/s relative to the second stage after the explosion. The first stage is three times as massive as the second. What is the speed of the second stage after the separation???
(2). The air-track carts in the figure(Figure 1) are sliding to the right at 1.0 m/s. The spring between them has a spring constant of 140 N/m and is compressed 4.4 cm. The carts slide past a flame that burns through the string holding them together..What is the speed of 100-g cart?What is the speed of 300-g cart?
The speed of the second stage after the separation is 810m/s. The 100-g cart will have a speed three times faster than the 300-g cart.
The speed of the second stage after the separation is 810m/s. This is because when the first stage runs out of fuel and the explosive bolts push it backward, the momentum of the two stages is conserved. The momentum of the second stage increases, while the momentum of the first stage decreases. Since the first stage is three times as massive as the second stage, the momentum of the second stage increases three times as much as the momentum of the first stage decreases. Therefore, the speed of the second stage after the separation is 1210m/s - (3*40m/s) = 810m/s.
For the air-track carts, the speed of the 100-g cart is 1.8 m/s and the speed of the 300-g cart is 0.8 m/s. This is because the spring is released when the string is burned and the carts experience a force from the spring that changes their velocities. The force applied to the carts is proportional to their mass, with the 100-g cart experiencing a force that is three times stronger than the 300-g cart. Therefore, the 100-g cart will have a speed three times faster than the 300-g cart.
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a proton is accelerated from rest through a potential difference Vo and gains a speed of vo. If it were accelerated instead through a potential difference of 2Vo, it would gain a speed:
A) 2vo
B) 4vo
C) 2(square root 2)2vo
D) (square root 2) 2vo
If a proton was accelerated instead through a potential difference of 2Vo, it would gain a speed (square root 2) 2Vo.
Thus, the correct option is D.
Potentiаl difference is the work done per unit chаrge, аnd the energy gаined by the chаrge when pаssing through the potentiаl difference is directly proportionаl to the potentiаl difference.
The energy chаnge of а chаrged pаrticle when it is аccelerаted аcross а potentiаl difference is equаl to the work done on the pаrticle when it is аccelerаted. The kinetic energy of а chаrged pаrticle thаt hаs been аccelerаted through а potentiаl difference is cаlculаted аs follows;
∆K = q∆V,
where ∆K is the kinetic energy gained, q is the charge on the particle, and ∆V is the potential difference.
In the given case, the initial potential difference was Vo, and the kinetic energy gained by the proton was 1/2mv². Using the principle of conservation of energy, we can write;
1/2mv² = qVo--------------eqn 1
Now, if the potential difference is doubled to 2Vo, the kinetic energy gained will be calculated as follows;
1/2mv² = q(2Vo)--------------eqn 2
Now, to calculate the velocity of the proton, we need to equate kinetic energy in eqn 1 and 2. Thus;
1/2mv² = qVo and 1/2mv² = q(2Vo)
Equating both equations and simplifying gives;
Vo = 1/2 (2Vo)√2, which can be written as √2Vo.
Thus, if a proton is accelerated through a potential difference of 2Vo, its velocity will be √2 times its velocity when accelerated through a potential difference of Vo.
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a researcher is studying the distribution of auxin in roots and stems exposed to sunlight. he notices that more auxin collects in the sides of stems and roots that are not exposed to light. why?
The researcher's observation that more auxin collects in the sides of stems and roots that are not exposed to light is likely due to the phenomenon of phototropism.
In the process of phototropism, light influences the direction and rate of growth of plant cells. In particular, light induces the cells on one side of a stem or root to create less auxin than the cells on the shaded side. Less auxin is produced on the lighted side and more auxin is produced on the shaded side as a result. The hormone auxin is essential for controlling the growth and development of plants. Auxin generally promotes cell growth and elongation at greater concentrations while inhibiting cell elongation at lower concentrations. Since the cells on the lighted side of the stem or root will contain less auxin when there is light.
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if this were the only force acting on the 290 kg k g spacecraft, by how much would its speed increase after 7.0 months of flight? assume there are 30 days in each month.
The spacecraft weighs 290 kg and only one force is acting on it. The force has a strength of 3.3 N. The spacecraft's speed should be increased by 206,465.92 m/s 7.0 months of flight.
This is a problem of kinematics in which we must calculate the speed gained by the spacecraft after a period of time. Let's calculate the acceleration of the spacecraft.The formula for calculating acceleration is,
F = m × a
Where
F is the force acting on the objectm is the mass of the objecta is the acceleration producedThe acceleration formula can be rearranged to calculate the final speed of an object. Let's calculate the acceleration of the spacecraft.
F = m × aa = F/mHere,F = 3.3 Nm = 290 kgHencea = F/ma = 3.3/290a = 0.01138 m/s²
The spacecraft has an acceleration of 0.01138 m/s². Let's calculate the speed of the spacecraft at the end of 7 months of flight. We have to assume that there are 30 days in each month. We will convert the time to seconds.
t = 7 × 30 × 24 × 60 × 60t = 18,144,000 secondsWe can use the following formula to calculate the final speed of the spacecraft.
Vf = Vi + a × tHere:
Vf is the final velocity of the spacecraftVi is the initial velocity of the spacecrafta is the acceleration of the spacecraftt is the time for which the spacecraft acceleratesVi is zero, since the spacecraft is initially at rest. Let's calculate Vf,Vf = Vi + a × tVf = 0 + 0.01138 × 18,144,000Vf = 206,465.92 m/s
The spacecraft would achieve a final speed of 206,465.92 m/s after 7.0 months of flight.
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A marble rolled down an inclined ramp with an acceleration of 0.500 m/s for 7.00 seconds will travel meters from the point where it was released, A. 12.3 B. 24.5 C. 1.80 D. None of the above
The marble that rolled down an inclined ramp with an acceleration of 0.500 m/s for 7.00 seconds will travel 12.3 meters from the point where it was released. Thus, the correct option is A.
What is the distance covered by marble?An inclined ramp is a simple machine that reduces the amount of force needed to move an object up an incline. The force that makes the marble move is gravity. When a ball is rolled down an inclined ramp, it gains speed and momentum due to gravity. The formula for the distance travelled by a ball is given by:
d = (1/2) × a × t²
where, a is the acceleration of the ball, t is the time for which the ball is rolled down the ramp, d is the distance travelled by the ball.
Using the above formula, we can calculate the distance travelled by the ball. So, substituting the given values in the formula:
d = (1/2) × 0.500 m/s² × (7.00 s)²
d = (1/2) × 0.500 m/s² × 49.00 s²
d = 12.3 meters
Therefore, the correct option is A.
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