A charged particle moves in a circular path in uniform magnetic field, then following would increase the period of the particle's motion: Increasing its charge and Increasing the field strength.
What would increase the period of the motion of particle?For charged particle moving in circular path in uniform magnetic field, centripetal force is provided by magnetic force,
q v B = mv²/r
Here, q is charge, v is velocity and B is magnetic field
m is mass, r is radius of the orbit
The period of the motion is given as :
T= 2πr/v
Substituting value of T in previous equation.
Now, q v B= m Tv³/2π
Hence, T= 2πq B/mv²
Here, we can clearly see that the period is: directly proportional to the charge and the magnetic field but inversely proportional to mass and square of the speed.
Therefore, increasing its charge and the field strength will increase the period of the particle's motion.
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two objects with equal masses are separated by a distance of 1 meter. if the mass of one of the objects were to be doubled, then the gravitational force between the two masses would be which of the following?
If the mass of one of the objects were to be doubled, then the gravitational force between the two masses would be 2 times greater than it was originally.
The gravitational force between two masses is given by the equation:
F = G x (m1 x m2) / r^2
where F is the gravitational force, G is the gravitational constant (equal to 6.67 x 10^-11 Nxm^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them. If the mass of one of the objects is doubled, the gravitational force between the two masses will also be doubled. This is because the gravitational force is directly proportional to the masses of the two objects. Therefore, if the mass of one of the objects is doubled, the gravitational force between the two masses will be 2 times greater than it was originally.
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11. in which direction are nearly all galaxies moving? 12. what is hubble's law? 13. match the terms with their definitions
Hubble's law is a statement of a direct correlation between the distance to a galaxy and its recessional velocity as determined by the red shift. It states that all galaxies are moving away from the Earth and from each other.
Edwin Hubble discovered that most of the galaxies are moving away from the Earth and away from each other. He also discovered that there is a relationship between the distance to a galaxy and its speed. He stated that the farther away a galaxy is, the faster it is moving away from us. Hence, Hubble’s law, also known as the Hubble–Lemaître law, refers to the observation in physical cosmology that galaxies are moving away from Earth at speeds proportional to their distance.
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5. a particular satellite with a mass of 200kg is put into orbit around ganymede (the largest moon of jupiter) at a distance 300 km from the surface. what is the gravitational force of attraction between the satellite and the moon? (ganymede has a mass of 1.48x1023 kg and a radius of 2631 km.) (4 pts)
The gravitational force of attraction between the satellite and the moon is 1.44 x 10^22 N.
To calculate the gravitational force of attraction between the satellite and the Ganymede moon, you can use the formula:
F = (G * m1 * m2) / r^2where F is the gravitational force, G is the gravitational constant (which is approximately 6.67 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
Plugging in the values provided, we get:
F = (6.67 x 10^-11 N*m^2/kg^2) * (200 kg) * (1.48 x 10^23 kg) / (300 km)^2
= 1.44 x 10^22 N
This is the gravitational force of attraction between the satellite and the Ganymede. It is a very strong force, due to the large mass of the moon and the relatively close distance between the satellite and the moon.
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You want exactly nine gallons of water, but you have only a seven-gallon bucket and a five-gallon bucket. Although you, at first, do not see how these two buckets can help, you realize that you can use them in the following way: fill the seven-gallon bucket, then pour it in the five-gallon bucket, leaving two gallons in the seven-gallon bucket. Pour out the five-gallon, pour the two gallons of water in the five-gallon and refill the seven-gallon bucket and you have a total of nine gallons of water. To solve this problem, you had to use
a. Selective comparison.
b. Selective attention.
c. Selective encoding.
d. Selective combination.
You want exactly nine gallons of water, but you have seven-gallon bucket and five-gallon bucket. To solve this problem, we use the method of selective combination.
We should fill the seven gallon bucket and then pour it into five gallon bucket, leaving the two gallon behind. Discard five gallons and pour two gallon into five-gallon bucket and refill seven-gallon bucket. You will have a total of nine gallons.
To elaborate, the selective combination is an insight producing process that occurs when one merges two or more components that are seemingly unrelated to produce a working solution.
In this case, two components combined are the five-gallon bucket and the seven gallon bucket.
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Give two reasons why the current concentration of carbon dioxide in the atmosphere is far higher than the natural range over the last 650, 000 years and is increasing rapidly.
The current concentration of carbon dioxide in the atmosphere is far higher than the natural range over the last 650,000 years, and is increasing rapidly for several reasons. Some of the main factors that have contributed to this increase include:
Burning of fossil fuels: The most significant source of carbon dioxide emissions is the burning of fossil fuels, such as coal, oil, and natural gas. These fuels are rich in carbon, and when they are burned, they release large amounts of carbon dioxide into the atmosphere. Over the past century, the global demand for energy has increased dramatically, leading to a corresponding increase in the burning of fossil fuels and the release of carbon dioxide into the atmosphere.Deforestation: Trees and other vegetation absorb carbon dioxide from the atmosphere as part of the process of photosynthesis. When trees are cut down and burned, or otherwise removed from the landscape, this carbon is released back into the atmosphere. Deforestation is therefore a major source of carbon dioxide emissions, and can contribute to the increasing concentration of carbon dioxide in the atmosphere.Overall, the current high levels of carbon dioxide in the atmosphere are the result of human activities, such as the burning of fossil fuels and deforestation, which have released large amounts of carbon dioxide into the atmosphere. These activities have led to a rapid increase in the concentration of carbon dioxide in the atmosphere, far beyond the natural range of concentrations over the last 650,000 years.
If the tank is a reusable tank, there are no check valves and system refrigerant can enter the tank as long as the system pressure is higher than the pressure in the tank. True or False
It is true that if the tank is a reusable tank, there are no check valves and system refrigerant can enter the tank as long as the system pressure is higher than the pressure in the tank.
A fundamental valve type dispersed generously all through a modern refrigeration framework is a manual shutoff valve. In the totally vacant position, this valve ought to permit a free progression of refrigerant and when shut totally block the stream. The standard capability of the shutoff valve is to confine a part or a segment of the framework.
Air conditioning Valves are such parts without which we have no control over stream in pipes .Valves are required for each medium whether it is water, gas, air or some other fluid. Valves would have been required for solids in the event that they could have capacity to stream. Like different regions, valves track down its broad use in air conditioning.
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TRUE OR FALSE When four-wheel drive is selected in a vehicle equipped with a four-wheel drive transfer case, the mode fork, which is similar to a shift fork found in a manual transmission, moves a synchronizer sleeve to the input shaft chain sprocket.
True
False
This statement is True because A synchronizer sleeve is moved to the input shaft chain sprocket by the mode fork, which resembles a shift fork in a manual transmission, once four-wheel drive .
Is manual transmission a stick shift?When a vehicle has a manual transmission, the driver controls the clutch and selects the appropriate gear. Most drivers believe a manual transmission, or stick gear as it is commonly known, gives them a greater sense of involvement with the car's operation and makes driving more enjoyable.
Why would you use a manual transmission?A manual gearbox is essentially a gear train that allows the driver to select from a variety of gear ratios to operate the vehicle. Higher gear ratios provide less torque but more speed, while lower gear ratios provide more torque but less speed.
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a string and allowed to revolve in a circle
of radius 1.7 m on a frictionless horizontal
surface. The other end of the string passes
through a hole in the center of the surface,
and a mass of 1.5 kg is tied to it, as shown.
The suspended mass remains in equilibrium
while the puck revolves on the surface.
0.029 kg
1.7 m
1.5 kg
What is the magnitude of the force that
maintains circular motion acting on the puck?
The acceleration due to gravity is 9.81 m/s?.
Answer in units of N.
part 2 of 2
What is the linear speed of the puck?
Answer in units of m/s.
The magnitude of the force that maintains circular motion acting on the puck is 15.68 Newtons and the linear speed of the puck is 30.854 m/s.
Uniform circular motion is the motion in a circle at constant speed and Instantaneous velocity is always tangent to the circle. The acceleration, called the centripetal acceleration, points toward the center of the circle.
For an object to be in uniform circular motion, Newton’s 2 nd law requires a net force acting on it. This net force is called centripetal force. Physically, the centripetal force can be the tension in a string, the gravity on a satellite, the normal force of a ring, etc.
Don’t count the centripetal force as an additional force in the free-body-diagram.. It refers to the required net force for circular motion.A centrifuge works by spinning very fast. An object in the tube requires a large centripetal force. When the liquid can’t provide such a large force, the object will move (sink) to the end of the tube.
we have,
Force = mg = 1.6 x 9.8 = 15.68 Newtons
now Force = F = m'v^2/R
15.68 = 0.028 x v^2/1.7
Speed = v = 30.854 m/s
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answer to this required:
Answer:
Explanation:
1) Conduction - 100% (C)
2) Convection can take place in vacuum is incorrect - 100% (A)
3) Remains same - 100% (D)
Throughout the reflection, make sure
you have a copy of the Student Guide and your data tables.
In this experiment, the
was intentionally manipulated. This was the independent variable
The dependent variable measured was the
Can someone please help with this physics problem.
Normal force that will act on the body moving downward will be 480 newtons.
What is Newton's second law?One of the most significant laws in all of physics is Newton's second law. F = ma, wherein F (force) as well as a (acceleration) both are vector values, can be used to represent an object with mass m is constant. A body accelerates according to the equation if there is a net force that acts on it.
what is Newton's first law?Until and unless an external force occurs on a body, it is in a condition of rest or homogeneous movement in a straight line. According to Newton's first law of motion, a body won't begin to move unless and until an outside force impacts it.
Briefing:given:
m=20 kg, a= 2.0m/s^2, g=10 m/s^2
Mg-F(normal)=Ma
60*10-F(normal)=60*2
F(normal)=480N
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a mass m at the end of a spring oscillates with a frequency of 0.92 hz . when an additional 800 g mass is added to m, the frequency is 0.58 hz .
The value of m is 0.534 kg
Given:f1 =0.92 Hz
f2 = 0.58 Hz
Δm = m2 - m1 = 800 g kg
Mass-spring oscillation occurs with a frequency of:
f = (1/2π)√(k/m)
k = m(2πf)²
When we have: we may conclude that k is constant for both trials.
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0. 800 g/ (0.92/0.58)^2 – 1
= 0.534 kg or 0.53kg or 534 g
Does a mass M vibrate when it is connected to a spring?A mass m linked to a spring produces an oscillation with a period of 4 s. A spring's time period lengthens by 1 s when a 2 kg additional mass is added to it.
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two balls of the same mass are thrown towards a wall and collide with it moving with a speed of 5 m/s. ball a hits the wall and rebounds with a speed of 4 m/s. ball b hits the wall and stops. assume that the collisions times are the same for each ball. compared to ball b, ball a has velocity change, momentum change, and impact force.
Two balls of the same mass are thrown towards a wall and collide with it moving with a speed of 5 m/s. Ball A hits the wall and rebounds with a speed of 4 m/s. Ball B hits the wall and stops. Assume that the collisions times are the same for each ball. Compared to ball B, ball A has a greater velocity change a greater change, and a greater impact force.
The term "impact force" refers to a circumstance in which effort is expended to move a target object a certain distance. When two objects collide, it can be understood as the force that results. The coming together of two objects is known as an object collision. A lot happens to an object as a result of the impact force quickly. F serves as a symbol to represent it. The dimensions are specified by [M1L1T-2] and the Newton (N) unit of measurement is used. Its equation is the twice-taken time divided by the product of a body's mass and velocity. In other words, it is the proportion of kinetic energy that a body has to its total distance traveled.
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in a clear, coherent paragraph-length response, indicate whether the acceleration of the system is greater in case a, case b, or the same for both cases and explain your reasoning. also indicate whether the force of contact that block m exerts on block 3m is greater in case a, case b or the same in both cases and explain your reasoning.
The (2) and (3), that contact force in case (a) is greater than in case (b).
What is force ?
Everyday activities like walking, setting something down on a surface, tossing something into the air, and even the tides' regular variations all include the application of force. The result of the interaction between two or more things, a force is a push or a pull.
What is acceleration ?
Acceleration is the rate at which velocity changes, as well as how quickly velocity changes over time. instantaneous acceleration is an acceleration at a single moment instantaneous acceleration divided by time deceleration
N1 and N2 are normal forces
T1 and T2 are contact force , (equal in magnitude, but opposite in direction)
i.e |T1| - |T2|
here for both case, acceleration
a = Fo/m1+ m2
a = Fo/M + 3M = Fo/um -(1)
acceleration for both cases remains same, as two blocks move with a single force fo, together
Fo – T2 = Ma or T1 = 3Ma
= T1 = 3M (Fo/4M)
T1 = 3/4 Fo –(2)
Fo – T2 = 3Ma
T2 = Ma
T2 = M (Fo/4) – (3)
It is obvious from (2) and (3), that contact force in case (a) is greater than in case (b).
Therefore, (2) and (3), that contact force in case (a) is greater than in case (b).
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a small sphere radious 0.05 surrounds the point (2, 3, -1). the flux of a vector g into this sphere 0.00004pi. estimate div g at the point (2, 3, -1).
The flux of a vector g into this sphere 0.00004pi. estimate div g at the point (2, 3, -1) is the divergence at the point is 0.03.
Calculation :
Divergence of a vector field G at a point p is defined as:
div(G)p=Fluxp/Volumep
A point p=(3,3,−1) has a radius of 0.1 units and the flux entering p is 0.00004π
.Volume of point p = (4/3)π 0.1³=0.0043π
Therefore, the divergence at the point is given by:
div(G)p=−0.00004π/(0.004/3)π=−0.03
divergence, in mathematics, the differential operator applied to a vector-valued function in three dimensions. The result is a function representing the rate of change. The divergence of vector v is given by . where v1, v2, and v3 are the vector components of v, typically the fluid flow velocity field.
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Communications satellites generally occupy a special orbit in which they go completely around the Earth once a day; since the Earth also turns once a day, the satellite is therefore always over the same point on the ground. Your satellite dish can then always be pointed at the same location in the sky.
A typical electromagnetic signal from a communications satellite might have an intensity of 0.750 pW/m2 at the surface of the Earth. Your satellite dish collects the energy of the electromagnetic signal from the satellite; then your dish passes that energy to its central receiver, where the signal can be processed by a circuit. The area of a satellite dish can be calculated from pi (3.1416) times the radius squared. If your satellite dish has a radius of 0.38 m, and if it is receiving the signal described above, how much electromagnetic energy is reflected to the central receiver in 3.5 minutes? Give your answer in picojoules.
In 3.5 minutes, the electromagnetic energy of the signal stated above, 71.45 pJ, is reflected to the main receiver.
Intensity = 0.75 x 10⁻¹² W/m²
Intensity = Power / Area
here, Area = A = π r = π (0.38)² = 0.4536 m²
so, Power = 0.75 x 10⁻¹² x 0.4536 = 3.402 x 10⁻¹³ W
Energy = Power x time
here, time = 3.5 minutes = 3.5 x 60 seconds
therefore, energy reflected = 3.402 x 10⁻¹³ x 3.5 x 60 = 71.45 x 10⁻¹² J = 71.45 pJ
this is the energy reflected the central receiver in 3.5 minutes.
Radiation with electromagnetic properties is electromagnetic energy. The energy connected to electromagnetic waves is commonly referred to by this phrase. Energy is what causes these waves to move through any medium. Photons, a unit of light energy that electromagnetic radiation travels in, are free of charge and mass. One of the most significant elements in the cosmos is electromagnetic energy. The force behind the electromagnetic waves is it.
When a force outside of the electrical charge accelerates it, electromagnetic energy is released. As a result of the acceleration, a wave of alternating magnetic and electric fields is created. This wave separates from the charge and flows independently through the medium.
The discrete energy packets known as photons make up electromagnetic waves.
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An inductor is connected to the terminals of a battery that has an emf of 10.0 V and negligible internal resistance. The current is 4.86 rmmA at 0.940 ms after the connection is completed. After a long time the current is 6.45 mA.
A)
What is the resistance R of the inductor?
B)
What is the inductance
The inductor's resistance, R, is 1550.38.
1.05 H is the inductance L.
What distinguishes an inductor from a capacitor?A capacitance opposes a variation in voltage, while an inductance opposes a changes in current. This is one of the key distinctions between the two components. Additionally, the capacitor and inductor both store energies with in type of either a magnetic charge and a magnetic force, respectively.
Briefing-
Given:
Emf, E=15.5V
Current, i = 4.86mA
Time, t = 0.940 ms
Maximum current, imax Calculation: 6.45mA
a) The resistance is given by,
Ꭱ= V/Imax = 10/6.45×10⁻³= 1550.38Ω
b) The inductance is given by,
L= -Rt/(1-i/imax)= 1550.38×0.940×10⁻³/In(1-4.86/6.45)
The inductance L is 1.05 H.
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PLS HELP NEED URGENT What is the value of g, the gravitational field strength, at the surface of the planet Venus?
Round your answer to two decimal places
Answer:
8.83 N/kg
Explanation:
To solve this problem we can use the formula:
[tex]g=G\frac{M}{r^{2} }[/tex] which describes the field strength at a certain distance from a body (where g=gravitational field strength, G=Gravitational constant, M=Mass of planet, and r=distance from center of planetary mass).
We are given the information that the radius of Venus is 6.073 x 10^6m and its mass is 4.88x10^24kg.
We are also given the gravitational constant.
Since the radius of a planet is equal to the distance between the center of the planet to its surface we can use this value for 'r'.
Therefore we now have all the information we need and so we can plug these values into the equation to solve for the gravitational field strength:
[tex]g=(6.67*10^{-11})\frac{(4.88*10^{24})}{(6.073*10^{6})^{2}}[/tex]
[tex]g=(6.67*10^{-11})\frac{(4.88*10^{24})}{3.688*10^{13}}[/tex]
[tex]g=(6.67*10^{-11})*(1.323*10^{11})[/tex]
[tex]g=8.83Nkg^{-1}[/tex] (to 2.d.p)
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 50.0 mm , and the potential difference between them is 345 V .
a. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
b. What is the magnitude of the force this field exerts on a particle with a charge of 2.00 nC ?
c. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
d. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
We are given big parallel metallic plates aa and bb bring contrary expenses of same significance are separated through a distance dd = 45.zero mm = zero.1/2 m wherein the capacity distinction among the 2 plates is V_V ab = 360 V
Required(a) The electric powered area among the 2 plates EE(b) The pressure FF exerted because of the electrical area.(c) The paintings performed WW(d) Compare the paintings performed fee to the extrade in capacity energy. Our trouble is associated with instance 23.9, wherein the electrical area among parallel plates is associated with the capacity distinction among the 2 plates through the following equationE = dfrac{V_}E= dV ab Where for a given capacity distinction V_V ab , the smaller the space dd among the 2 plates, the more the significance EE of the electric area.Now we are able to plug our values for V_V ab and dd into equation (1) to get EE begin E &= dfrac{V_} &= dfrac}{zero.1/2 ,text} &= eight instances 10^ ,text &= boxed{eight instances 10^ ,text} endE = dV ab = zero.045m360V =eight×10 3 V/m= eight×10 3 N/C Result2 of 2(a) E = eight.zero instances 10^ ,textE=eight.zero×10 3 N/CRead more about energy:
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What is the tension in the string in (Figure 1) ? The volume of plastic ball is 75 cm3 and the density is 840 kg/m3. Express your answer to two sig fig and include appropriate units.
The tension in the string is = 0.118
Given the values in the question,
The volume of the plastic ball = 75 [tex]cm^{3}[/tex]
Density ( ρ ) = 840 kg / [tex]m^{3}[/tex]
Let the tension on the ball = T
Since the density is given in centimeters so convert volume also in meters,
⇒ 75 [tex]cm^{3}[/tex] = 0.000075 [tex]m^{3}[/tex]
The ball is following the principle of buoyant force, so the ball is stable in the water the force that the ball is exerting into the water is equal to the force the water is exerting on the ball.
The forces could be represented as -
∑ [tex]F_{y}[/tex] = [tex]F_{b}[/tex] - T - [tex]W_{b}[/tex] = 0
T = [tex]F_{b}[/tex] - [tex]W_{b}[/tex]
T = ρ x g x [tex]V_{w}[/tex] - ρ x g x [tex]V_{b}[/tex]
T = ( 1000 x 9.8 x 0.000075 ) - ( 840 x 9.8 x 0.000075 )
T = 0.735 - 0.617
T = 0.118
Therefore, The tension in the string is = 0.118
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the total mechanical energy of the system is the same when the ball is at the top and bottom of the vertical circle. use conservation of energy to write an expression for v2t in terms of v2b . your answer may also include m , g , and l .
The expression for [tex]v^{2}_t[/tex] is equal to √5 times of [tex]v^{2}_b[/tex] where [tex]v_t[/tex] and [tex]v_b[/tex] are the velocities of ball at top and bottom of the circle respectively.
Consider the centripetal force at any point.
T – mg cosθ = m v² ⁄ r
T = mg cosθ + m v² ⁄ r
From figure, cosθ = (r – h) ⁄ r
Substitute in the equation of T.
T = m g (r – h) ⁄ r + m v² ⁄ r
= m ⁄ r (g (r – h) + v²)
But v² = u² – 2 g h
T = m ⁄ r (g r – g h + u² – 2 g h)
= m ⁄ r (u² – 3 g h + g r)
This is an equation for tension in string.
Now, when ball is at the top of circle
We have,
v = √(u² – 2 g h)
= √(5 g r – 2 g ×(2 r))
= √(5 g r – 4 g r)
= √(g r)
And, when ball is at the bottom of the circle,
T[tex]_H[/tex] > 0
m ⁄ r ×(u² – 5 g r) > 0
u² – 5 g r > 0
u² > 5 g r
u > √(5 g r)
Hence,[tex]v^{2}_t[/tex] is equal to √5 times of [tex]v^{2}_b[/tex]
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for lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of italian ham. the slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 n/m . the slices of ham are dropped on the plate all at the same time from a height of 0.250 m . they make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (shm). you may assume that the collision time is extremely small. Express your answer numerically in meters and take free-fall acceleration to be g= 9.80m/s^2
What is the period of oscillation T of the scale?
Express your answer numerically in seconds.
The period of oscillation T of the scale is T = 0.37 sec .
In the question ,
it is given that ,
weight of the Italian ham (m₁) = 0.300 Kg = 0.3 Kg
weight of the slices of ham are (m₂) = 0.400 kg = 0.4 Kg
the force constant of the vertical spring is(k) = 200 N/m .
Let oscillation period be = T ,
the plates make a totally inelastic collision ,
We know that , In inelastic collision, momentum is conserved but the kinetic energy is not conserved.
So , when the slice of ham land on the plate, then the kinetic energy of system equal to maximum potential energy of spring.
that means , T = 2π√(m₁ + m₂)/k
Substituting the values ,
we get ,
T = 2π√(0.3 + 0.4)/200
Simplifying further ,
we get ,
T = 0.37 seconds .
Therefore , If the collision is inelastic then the Period of oscillation is 0.37 s .
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Calculate the wavelengths of the components of the first line of the Lyman series, taking the fine structure of the 2p level into account. (Give your answers to at least four decimal places. Use values hc = 1239.8 eV · nm and E1 = 13.606 eV.)
smaller value: ____ nm
larger value: ____ nm
I already asked this but received an incorrect answer.
The smaller wavelength is 121.49515 nm and the larger wavelength is 121.4956866 nm.
It is the series of spectral lines which results in emission of UV radiations.
It happens when electron goes from n≥ 2 to n = 2
where, n = principle quantum number
The first line of Lyman series occur due to transition of electrons from n=1 to n=2
If we consider the '2p' level
The first transition of Lyman series
Energy:
-3.4015 eV ; n=2
-13.606eV ; n=1
ΔE = 4.5 * 10⁻⁵eV
Given
hc = 1239.8 eV(nm)
We know wavelength of radiation emitted is
\lambda = hc / (E₂-E₁)
If e⁻ transition occurs from n = 2P(3/2) to n = 1
E₂ - E₁ = -3.4015 eV + ΔE/2 + 13.606 eV
E₂ - E₁ = 10.2045225 eV
If e⁻ transition occurs from n = 2P(1/2) to n = 1
E₂ - E₁ = -3.4015 eV + ΔE/2 + 13.606 eV
E₂ - E₁ = 10.2044775 eV
Smaller value of wavelength = hc/E₂-E₁ = 1239.8/10.2045225 = 121.49515 nm
Larger value of wavelength = hc/E₂-E₁ = 1239.8 / 10.2044775 = 121.4956866 nm
Therefore the smaller wavelength is 121.49515 nm and the larger wavelength is 121.4956866 nm.
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In the original Bose-Einstein experiment, rubidium atoms were trapped in a small (but macroscopic) device, roughly 0.1 mm on a side. Consider a single Rb atom trapped at t0. If released (from rest), what happens to this particle as time goes by? What are m, a, and Ω here, as numbers? What is the "timescale" in seconds?
Think about a single Rb atom held at t0. if allowed (from rest). See McIntyre 6.40 for a hint. There, the solution is (x, t) = (2p) TTT, with 9-2h/m. (You must determine dimensionless constant .)
The use of rubidium in Bose-Einstein condensate is explained?Rubidium-87 has a positive scattering length at low temperatures, making it mutually repulsive despite rubidium-85's greater abundance. This prevents all condensates bar the tiniest ones from collapsing.
Why is rubidium such a vital metal?As a getter, rubidium is a substance that mixes with and extracts trace gases from vacuum tubes. Additionally, it is utilized in the creation of specialty glasses and photocells. It is easily ionized,
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which statement about saturn's rings is not true? which statement about saturn's rings is not true? the rings must look much the same today as they did shortly after saturn formed. some features of the rings are shaped by small moons that actually orbit within the ring system. the rings are so thin that they essentially disappear from view when seen edge-on. the large gap known as the cassini division is shaped by an orbital resonance with the moon mimas, which orbits well outside the rings.
Statement about saturn's rings is not true is that the rings must look much the same today as they did shortly after saturn formed.
Saturn, the most beautiful planet in our solar system, is famous for its dazzling rings. These rings extend far into space and engulf many of Saturn’s moons. The brightest rings, visible from Earth in a small telescope, include the D, C and B rings, Cassini’s Division.
The A ring. Just outside the A ring is the narrow F ring, shepherded by tiny moons, Pandora and Prometheus. Beyond that are two much fainter rings named G and E. Saturn's diffuse E ring is the largest planetary ring in our solar system, extending from Mimas' orbit to Titan's orbit, about 1 million kilometers.
The particles in Saturn's rings are composed primarily of water ice and range in size from microns to tens of meters. The rings show a tremendous amount of structure on all scales. Some of this structure is related to gravitational interactions with Saturn's many moons, but much of it remains unexplained.
One moonlet, Pan, actually orbits inside the A ring in a 330-kilometer-wide (200-mile) gap called the Encke Gap. The main rings (A, B and C) are less than 100 meters (300 feet) thick in most places. The main rings are much younger than the age of the solar system, perhaps only a few hundred million years old.
Hence, the correct option is A.
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2. What is reflection of light?
ection is when light waves curve around objects.
ection is when light waves bend a surface.
lection is when light waves bounce off a surface. lection is when light waves travel through objects.
Reflection of light is when light waves bounce off a surface.
How does light reflection occur?Reflection occurs when light traveling through one material bounces off a different material. The reflected light continues to travel in a straight line, but in a different direction.
The incident light ray that land on the surface is called the reflected off the surface while the ray that bounces back is called the reflected ray.
For example, if an individual looks at a bird, light has reflected off that bird and traveled in nearly all directions. If some of that light enters the individual's eyes, it hits the retina at the back of the eyes. An electrical signal is passed to the person's brain, and the brain interprets the signals as an image.
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the moment of inertia is group of answer choices inversely proportional to the square of the angular velocity inversely proportional to the kinetic energy not dependent on the mass of the rotating object.
Inversely proportional to the rectangular of the angular velocity inversely proportional to kinetic energy no longer depends on the mass of the rotating item.
The kinetic energy of a rotating body is given as
K=1/ 2 Iω
The moment of inertia relies upon the distribution of mass.
subsequently, the kinetic energy relies upon each distribution of mass and angular pace.
Kinetic energy is the energy an object has due to its movement. If we want to accelerate an object, then we need to apply a force applying a force calls for us to do paintings. After paintings have been accomplished, energy has been transferred to the object, and the object might be shifting at a new steady pace.
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mcdougal products is considering the purchase of new equipment to place in its factory. the equipment would cost $365,000, have a ten-year useful life and a salvage value at the end of its useful life of $65,000. the company estimates that annual revenues and expenses associated with the equipment would be as follows: the payback period of the new equipment is closest to:
Form the given annual revenue and expenses the payback period of the new equipment is closest to 4 years .
What is Payback Period ?
Payback Period is defined as amount of time the company will have to wait before it recovers its investment.
the net annual cash flow can be calculated using the formula .
Net annual cash flow = (Net income) + (Non-cash items (i.e. depreciation))
From the table we get , net income = $60000
depreciation = $30000
So ,
Annual Cash Flow = 60000 + 30000 = $90000
We know that , Payback period is = (Initial investment)/(Annual cash flow)
= 365000/90000
= 4.055555 years
≈ 4 years .
Therefore , The payback period of the new equipment is closest to 4 years .
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8x62.5the pressure at sea level is 111 atmosphere and increases at a constant rate as depth increases. when sydney dives to a depth of 232323 meters, the pressure around her is 3.33.33, point, 3 atmospheres. the pressure ppp in atmospheres is a function of xxx, the depth in meters.
The answer is, p(x) = 0.1x + 1 (p(x) is the pressure at a depth x)
Just what is pressure?
The amount of force applied to a certain region is referred to as pressure.
1 atmosphere is the pressure at sea level.
Sydney dove to a depth of 23 meters.
23-meter depth equals 3.3 atmospheres of pressure
23-meter deep pressure equals pressure at sea level plus k.
Sydney dove to a depth of 3.3 meters, where 3.3 = 23 k + 1 is the diving rate constant.
By deducting 1 from either side, we arrive at 3.3 - 1 = 23k + 1 -1.
2.3 = 23k is obtained by eliminating the +1 and -1 from the right side.
After subtracting the 23s from the top and bottom of the left side, we arrive at k = 0.1 by dividing both sides by 23.
When we enter k's value into the equation, we obtain
Pressure at a specific depth is equal to sea level pressure plus k. * Sydney's depth dove
p(x) = 1 + k(x) [where p(x) is the pressure at a depth x]
=> p(x) = 1 + 0.1(x)
=> p(x) = 0.1x + 1
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explain the effect of the added mass on the pressure and the relationship between the fluid levels on the two sides.
Pressure and mass flow relationship It follows that increasing the pressure supplied to the intake section will increase the pressure differential between the inlet and outlet valves.
More people will attempt to hurry past the segment as a result. As a result, we may state that pressure and mass flow rate are directly related (gradient). Mass flow rate is the amount of a substance that moves per unit of time in physics and engineering. In SI units, it is measured in kilograms per second, and in pounds or slugs per second in US customary units. Although occasionally (Greek lowercase mu) is used, the typical symbol is (, pronounced "m-dot"). The force that is delivered perpendicular to an object's surface and expressed as a symbol (p or P) is known as pressure.
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