For each value of the principal quantum number n, what are the possible values of the electron spin quantum number m_s?

a. 0
b. -3/2
c. +3/2
d. +1/2
e. -1/2

Answers

Answer 1

Answer:

d) and e)

Explanation:

For any value of the principal quantum n, the only possible values of the electron  spin quantum number m_s are + 1/2 and -1/2.Since this number is related with the angular momentum of the electron, it has a magnitude (1/2) and an orientation given by the sign (+ or -)So, the only right answers are d) and e).

Related Questions

Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan

Answers

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?

a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2

Answers

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s²          { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

A physicist wants to estimate the rate of emissions of alpha particles from a certain source. He counts 400 emissions in 80 seconds. Estimate the rate, in emissions per second and find the uncertainty in this estimate.

Answers

Answer:

Emissions per second = 0.36

Explanation:

Please find the attached question

Solution

Given  

Let X be the rate of background emission.

X = B/t  

Where B = 36

And t = 100

X = 36/100 = 0.36

Determine the tension in the string that connects M2 and M3.

Answers

therefore mass m1=4.8 kg and the tension in the horizontal spring T2=10N.

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A golf ball is dropped from rest from a height of 8.40 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Answers

Answer:

t1= 8.40/10 =.84 s

t2 = 5.60/10 = .56s

t3= 1.4/10 = .14s

total time = 1.54 sec

A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answers

1.244 m per second the person driving will go

Use the following information for questions 18 - 21. A 0.13 kg disk is rotating at an angular speed of 57 rad/s. The disk has a radius of 0.25 m. The disk speeds up for 3 s. After the 3 s have passed, the edge of the disk is under a centripetal force of 312.13 N. What is the centripetal acceleration of the disk at this time

Answers

Answer:

[tex]a=2401m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.13kg[/tex]

Angular speed [tex]\omega=57rads/s[/tex]

Radius [tex]r=0.25m[/tex]

Time [tex]t=3s[/tex]

[tex]F=312.13N[/tex]

Generally the equation for centripetal acceleration is mathematically given by

[tex]a=\frac{f}{m}[/tex]

[tex]a=\frac{312.13}{0.13}[/tex]

[tex]a=2401m/s^2[/tex]

Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above​

Answers

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.

Answers

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]

The work done by gravity is:

[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]

Electromagnetic waves from the sun carry what to the earth

Answers

Answer:

Solar radiation

Explanation:

Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.

==>  Energy

==>  Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays

==>  They carry all these kinds of energy wherever they go.  Not only to the Earth.

A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the object for the next 4.00 s. How far has the object moved during this 7.00 s interval?

Answers

Answer:

The total distance at 7 s is:

[tex]x_{tot}=27\: m[/tex]

Explanation:

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_{1}=0.5at_{1}^{2}[/tex]

[tex]x_{1}=0.5(1.8)(3)^{2}[/tex]

[tex]x_{1}=8.1\: m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s[/tex]

So the second distance will be:

[tex]x_{2}=vt_{2}=5.4*4=21.6\: m[/tex]

Therefore, the total distance is:

[tex]x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m[/tex]

I hope it helps you!

The total distance at 7 s is: [tex]x_{total}=27\ m[/tex]

What is force?

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}=\dfrac{9}{5}=1.8\ \frac{m}{s^2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_1=0.5at_1^2[/tex]

[tex]x_1=0.5(1.8)(3^2)[/tex]

[tex]x_1=8.1\ m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_i+at_1=0+(1.8)(3)=5.4 \frac{m}{s}[/tex]

So the second distance will be:

[tex]x_2=vt_2=5.4\times 4=21.6\ m[/tex]

Therefore, the total distance is: [tex]x_{total}=27\ m[/tex]

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Four 8.5-kg spheres are located at the corners of a square of side 0.90 m.

A. Calculate the magnitude of the gravitational force exerted on one sphere by the other three.

B. Calculate the direction of the gravitational force exerted on one sphere by the other three.

Answers

Explanation:

the answer is in the above image

An object of mass 15 kg is at an elevation of 100 m relative to the surface of the Earth. What is the potential energy of the object, in kJ? If the object were initially at rest, to what velocity, in m/s, would you have to accelerate it for the kinetic energy to have the same value as the potential energy you calculated above? The acceleration of gravity is 9.8 m/s2.

Answers

Answer:

The potential energy of the object is 14.7 kJ

The velocity of the object is 44.27 m/s

Explanation:

Given;

mass of the given object, m = 15 kg

position of the object, h = 100 m

acceleration due to gravity, g = 9.8 m/s²

The potential energy of the object is calculated as;

P.E = mgh

P.E = 15 x 9.8 x 100

P.E = 14,700 J = 14.7 kJ

The velocity of the object if its kinetic energy must equal potential energy;

¹/₂mv² = mgh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 100)

v = 44.27 m/s

A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

Answers

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Two insects are at points 1 and 2. which
of the two insects has the greater kinetic energy of rotation

Answers

the second one because it has a longer way that the rotation has to go around. the first one will just be quicker. it depends how the question wants you to describe kinetic energy! if it wants quicker, then 1. if longer movement, then 2!

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7,calculate the maximum kinetic energy

Answers

Complete question:

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.

Answer:

the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹ J

Explanation:

Given;

work function, Ф = 1.8 eV

wavelength of the light, λ = 3 x 10⁻⁷ m

The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.

[tex]E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J[/tex]

[tex]K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J[/tex]

Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

what is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s​

Answers

Answer:

by using p = mv equation we can find v,

6 = 1.3 v

4.615 = v

If the frequency of a sound wave is 250 hertz and its wavelength is 1.36 meters, what is the wave's velocity
OA. 250 meters/second
ОВ.
340 meters/second
O c.
200 meters/second
OD.
120 meters/second

Answers

Answer:

B

Explanation:

V=frequency*wavelength V=? W=1.36mF=250hertz

Answer:

B

Explanation:

Bc i say

A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?

Answers

Vs = 34m/s
I don’t have an explanation my apologies.

When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

What is the Doppler formula?

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

Learn more about Doppler equation.

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13. What type of lens bends light outwards and away from a point?
concave

Answers

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.

Convex lens is the answer.

See the attached diagram.

what are two injuries that heal slowly and with difficulty

Answers

Wounds or sores that take more than a few weeks to heal might be infected and require medical treatment, and often indicate an underlying disease such as diabetes.

When you cut or burn yourself, your body begins a three-stage process to repair the damaged skin.

First, an immune response causes the wound to become inflamed to prevent infections.

Second, new cells (a scab) form over the wound, and finally scar tissue forms to heal the wound.

Some wounds heal easily while others can take longer, particularly if they are severe or the individual has a poor state of health.

g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is

Answers

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

     

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

16. What type of lens bends light inwards?
concave
convex

Answers

Answer:

convex lens because it makes parallel light rays passing through it bends inwards and meets convex at a spot just beyond the lens known as the focal point

Answer:

Explanation:

What is the law of conservation of energy ​

Answers

energy can not be created nor destroyed
Hope this helps!!

Water at 200 C has a bulk modulus of 2.2109 Pa, and the speed of sound in water at this temperature is 1480m/s. For 1000Hz sound waves in water at 200 C, what displacement amplitude is produced if the pressure amplitude is 310-2 Pa?​

Answers

What does the mystery stand for?

6. Show that the weight of an object on the moon is 1/6 its weight on earth.​

Answers

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat

Answers

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

[tex]Mv+mu=Mv_1+mv_2[/tex]

(900 x 47) + (200 x -30)  = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

36300 =  (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)

[tex]9v_1 = 363 - 2v_2[/tex]

[tex]v_1=\frac{363 - 2v_2}{9}[/tex]

The mathematical expression for the conservation of kinetic energy is

[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]

[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex]    ................(ii)

[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]  

[tex]21681 = 9v_1^2+2v_2^2[/tex]

Substituting (i) in (ii)

[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]

[tex](363-2v_2)^2+18v_2^2=195129[/tex]

[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]

[tex]22v_2^2-145v_2-63360=0[/tex]

Solving the equation, we get

[tex]v_2=96 \ m/s, -30 \ m/s[/tex]

The negative velocity is neglected.

Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get

[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]

     = 19

Thus, only impulse of importance is used to find final velocity.

A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping

Answers

Explanation:

The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

[tex]y:\:\:\:\:N - mg = 0[/tex]

[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]

or

[tex]m \dfrac{v^2}{r} = \mu mg[/tex]

Solving for [tex]\mu[/tex],

[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]

Impervious surfaces prevent water from flowing through them.
True or false?

Answers

True
Explanation: impervious surfaces are any materials that do not allow water to flow through them such a concrete and asphalt

Answer: the answer is true

Explanation:

Hi there! I'm not quite sure on how to solve this....

Answers

[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]

Explanation:

The volume of a cube is V = x^3. Taking the time derivative of this expression, we get

[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]

or

[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]

We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is

[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]

Answer:

Explanation:

[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]

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