a proton is in a box of width L. what must the width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound

Answers

Answer 1

Answer:

Width of a box, [tex]l=6.41\times 10^{-15}\ m[/tex]

Explanation:

The ground level energy of a proton in a box is, E = 5 MeV

[tex]E =5\times 10^6\ eV\\\\=5\times 10^6\times 1.6\times 10^{-19}\\\\=8\times 10^{-13}\ J[/tex]

Energy in a box is given by :

[tex]E=\dfrac{n^2h^2}{8ml^2}[/tex]

For ground state, n = 1

m is mass of proton

h is Planck's constant

l is width of the box

[tex]l^2=\dfrac{n^2h^2}{8mE}\\\\l^2=\dfrac{1^2\times (6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times 8\times 10^{-13}}\\\\l=\sqrt{\dfrac{1^{2}\times(6.63\times10^{-34})^{2}}{8\times1.67\times10^{-27}\times8\times10^{-13}}}\\\\l=6.41\times 10^{-15}\ m[/tex]

So, the width of the bx is [tex]6.41\times 10^{-15}\ m[/tex].

Answer 2

The width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.

What is the energy in a box?

The energy in a box can be calculated with the following formula.

[tex]E=\dfrac{n^2h^2}{8mL^2}[/tex]

Here, (E) is the energy at the nth state, (n) n is the quantum number, (h) is plank's constant and (L) is the width of the box.

The proton is in a box of width L. The width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound.

The energy of the box is at ground level. Then the value of nth state will be 1. It is known that the value of plank's constant is 6.63×10⁻³⁴ m²kg/s.

Put this values in the above formula as,

[tex]8\times10^{-13}=\dfrac{(1)^2(6.63\times10^{-34})^2}{8(1.67\times10^{27})(l)^2}\\l=6.41\times10^{-15}\rm\; m[/tex]

Thus, the width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.

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Related Questions

How original were Newton’s contributions to science? (In what ways did Newton depend on the mechanical view?)

Answers

Newton's new “reflecting telescope” was more powerful than ... Before Newton, scientists primarily adhered to ancient theories on ... laws of motion laid the groundwork for classical mechanics. Newton's research on motion helped give credibility to the heliocentric view. Newton also helped pioneer telescopic innovations, and he is sometimes credited with inventing the first reflecting telescope. He also conducted experiments using the prism, and developed a theory about the nature of color and light.

Compare the weight of a 60 kg person on the earth with the weight of the same person on
the moon. Then, describe a quick (but very costly) way for dieters
at NASA to lose weight.

Answers

Answer:

Explanation:

The formula for weight is

W = mg, where

W = the weight of the object or person

m = mass of the object or person

g = acceleration due to gravity

Now, we're given the mass of the person to be 60 kd, and thus, the weight of that person would be

W = 60 * 9.81

W = 588.6 N

On the surface of the moon, the weight of the person would be

W = 60 * 1.625

W = 97.5 N

Therefore, the weight of the person on both surfaces are 588.6 and 97.5 respectively

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg. The spring is released and the object moves along a frictionless surface when it reaches a small embankment. If the speed of the object is 2.45 m/s at location A, what is the embankment height h

Answers

Given :

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.

To Find :

The embankment in the height.

Solution :

Since no external force is acting in the system, therefore total energy will be conserved.

Initial kinetic energy of the object = Energy stored in spring

[tex]K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J[/tex]

Also, initial potential energy is 0.

Now,

[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m[/tex]

Therefore, the embankment height is 0.64 m.

a box is 30 cm wide, 40 cm long and 25 cm high calculate: what is the area of its base.​

Answers

Answer:

1200cm²

Explanation:

Width= 30cm

Length= 40cm

Area of base(A) = Width×Length

= 30cm×40cm

= 1200cm²

Taking into account the definition of reactanguar prism and area of rectangle, the area of its base is 1200 cm².

A rectangular prism is a polyhedron whose surface is formed by two equal and parallel rectangles called bases and by four lateral faces that are also parallel and equal rectangles two by two.

So, the base being a rectangle, its area is calculated as the multiplication between the base and the height. In this case, these values ​​correspond to the width and length of the figure.

Then, in this case, you know:

Width= 30cm Length= 40cm

Being:

Area of base (A) = Width×Length

Then, the are of base (A) is calculated as:

Area of base (A)= 30cm×40cm

Solving:

Area of base (A)= 1200cm²

Finally, the area of its base is 1200 cm².

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be when it is very far from the Earth

Answers

Answer:

The value is  [tex]v = 2.3359 *10^{4} \ m/s[/tex]

Explanation:

From the question we are told that

  The  initial speed is [tex]u = 2.05 *10^{4} \ m/s[/tex]

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             [tex]T__{E}} = KE__{i}} + KE__{e}}[/tex]

Here  [tex]KE_i[/tex] is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          [tex]KE_i = \frac{1}{2} * m * u^2[/tex]

=>       [tex]KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2[/tex]

=>       [tex]KE_i = 2.101 *10^{8} \ \ m \ \ J[/tex]

And  [tex]KE_e[/tex] is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       [tex]KE_e = \frac{1}{2} * m * v_e^2[/tex]

Here [tex]v_e[/tex] is the escape velocity from earth which has a value [tex]v_e = 11.2 *10^{3} \ m/s[/tex]

=>    [tex]KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2[/tex]

=>    [tex]KE_e = 6.272 *10^{7} \ \ m \ \ J[/tex]

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        [tex]KE_p = \frac{1}{2} * m * v^2[/tex]

Generally from the law energy conservation we have that

        [tex]T__{E}} = KE_p[/tex]

So

       [tex]2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2[/tex]

=>     [tex]5.4564 *10^{8} = v^2[/tex]

=>     [tex]v = \sqrt{5.4564 *10^{8}}[/tex]

=>     [tex]v = 2.3359 *10^{4} \ m/s[/tex]

g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that the average wavelength of each photon is 550 nm.

Answers

Answer:

The value is [tex]N = 1.107 *10^{45 } \ photons[/tex]    

Explanation:

From the question we are told that

   The  power output from the sun is  [tex]P_o = 4 * 10^{26} \ W[/tex]

   The average wavelength of each photon is  [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]

Generally the energy of each photon emitted is mathematically represented as

        [tex]E_c = \frac{h * c }{ \lambda }[/tex]

Here  h is the Plank's constant with value  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

          c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

       [tex]E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }[/tex]          

=>   [tex]E_c = 3.614 *10^{-19} \ J[/tex]          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

         [tex]N = \frac{P }{E_c}[/tex]

=>      [tex]N = \frac{4 * 10^{26} }{3.614 *10^{-19}}[/tex]

=>      [tex]N = 1.107 *10^{45 } \ photons[/tex]    

A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill then how much force did they exert?

Answers

Explanation:

ans is equal to 504j* 23 m* 10 ms


Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From
the frame of reference of car 1, what is the velocity of car 2?

Answers

Answer:

10 mph faster than car 1 is going

Explanation:

Ist Law: a object continues in a state of
rest or motion unless an external force
applied to it


This image shows an example of
the first law because:

Answers

Answer:

Explanation:

This image is the example of Newton's first law of motion because:

As per Newton's first law, football will remain in the state of rest until a player applies an external force by kicking the ball.

And the ball will keep on moving until unless the net of a goal post exerts the external force to stop the ball.

Two workers are sliding 500 kg crate across the floor. One worker pushes forward on the crate with a force of 440 N while the other pulls in the same direction with a force of 340 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

0.159

Explanation:

Given that

F1 = 440 N

F2 = 340 N

m = 500 kg

acceleration, a = 0 m/s²

Remember that F = m.a

If F - f(f) = 0, then F = f(f)

N = mg

f(f) = μN, substituting for N, we have

f(f) = μmg

The total force, F = F1 + F2

F = 440 + 340

F = 780 N

Recall that we'd already proven that

F = f(f), so F = 780 N = f(f)

And again, f(f) = μmg, if we substitute for all the values, we have

780 = μ * 500 * 9.81

780 = μ * 4905

μ = 780 / 4905

μ = 0.159

Therefore, the coefficient of static friction of the crate on the floor is 0.159

It is difficult to lift a bigger stone than the smaller
stone. why?
plzzz give me short and brillient answer​

Answers

A bigger object is harder to lift because it has more mass which means a higher weight. But not everything has the weight you’d assume, for example: an eraser is smaller than a paper but the eraser is still heavier.

What determines the type of air mass that forms in an area?
Question 11 options:
The amount of oxygen present
The amount of air present
The direction of air flow
The location where it forms

Answers

Answer: no because you have left the number

Explanation:

A toy car weighing 3.2 N gets pushed a distance of 10 m in 1.3 s. What is the momentum of the car?

Answers

Answer:

2.51 kg * m/s

Explanation:

In order to find momentum, use the equation below:

momentum = mass * velocity.

Since neither mass nor velocity was given, you must solve for both variables.

In order to solve for mass, use the force equation for its weight / gravitational force.

Fg (gravitational force) = 3.2 N = ma = 9.8m

mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg

In order to solve for velocity, use the equation:

velocity = displacement / time

velocity = 10m / 1.3 s = 7.69231 m/s

Momentum = mass * velocity = 0.326531 kg *  7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s

the tighter the threads of the screw the greater

Answers

That ain’t a question but okay
bru that wasnt a question

To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for the glove to reach its maximum height

Answers

Hello!!

For the maximum height the final velocity is zero, because can't up more.

Then, use the formula:

V = Vi + gt

Replacing, we have:

0 m/s = 5,3 m/s + (-9,8 m/s² * t)

0 m/s - 5,3 m/s = -9,8 m/s² * t

(-5,3 m/s) / -9,8 m/s² = t

t = 0,54 s

The time it will take to reach the maximum height is 0,54 seconds.

In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones are 0.60 m long and their mass is 4.0 kg. From wrist to the tip of the fingers, the hand bones are 0.10 m long and their mass is 1.0 kg. In the model above, each bone is assumed to have a uniform density. When the arm and hand hang straight down, the distance from the shoulder to the center of mass of the arm-hand system is most nearly

Answers

Answer:

0.37 m

Explanation:

Let the shoulder be the origin.

The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.

The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones  from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m

Substituting these into the equation for the center of mass, we have

y = (m₁y₁ + m₂y₂)/(m₁ + m₂)

y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)

y = (1.20 kgm + 0.65 kgm)/5.0 kg

y = 1.85 kgm/5.0 kg

y =  0.37 m

The distance of the center of mass from the shoulder is thus y = 0.37 m

What happens between particles of different charges?

Answers

Answer:

Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other.

Explanation:

oo hi granger ru online here i have a doubt in physics . ​

Answers

Answer:

.

Explanation:

The temperature of the surface of the Sun is 5500°C.

a. What is the average kinetic energy, in joules, of hydrogen atoms on the surface of the Sun?
b. What is the average kinetic energy, in joules, of helium atoms in a region of the solar corona where the temperature is 6.00 times 10^5 K?

Answers

Answer:

a. the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. the average kinetic energy of helium atoms is 1.24 × 10^-17J

Explanation:

The computation is shown below;

As we know that

Kinetic energy = 3 ÷ 2 kT

where,

K = Boltzmann constant

And, T = Temperature

a. Now the temperature in kelvin is

T = (5,500 × (°C ÷ K) + 273.15 K)

= 5773.15 K

As

Kinetic energy = 3 ÷ 2 kT

So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)

= 1.20 × 10^-19J

hence, the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. As

Kinetic energy = 3 ÷ 2 kT

now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )

= 1.24 × 10^-17J

hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J

18
AX
A bicycle rider travels 15 km in 1.25 hours. What is the rider's average speed? v=
At
rage speed [v-ante
10.5 km/h
13.75 km/h
12 km/h
22.5 km/h

Answers

Average speed= total distance/total time =12km/h

The required average speed of bicycle rider is 12 km/h.

Given data:

The distance travelled by the bicycle rider is, d = 15 km.

Time taken to cover the distance is, t = 1.25 h.

Here, we need to use the simple relation between the speed, time and distance. The distance covered covered by any object per unit of time is known as average speed of object. And the mathematical expression for the average speed is given as,

[tex]v = \dfrac{d}{t}[/tex]

Here, v is the average speed.

Solving as,

[tex]v = \dfrac{15}{1.25}\\\\v = 12\;\rm km/h[/tex]

Thus, we can conclude that the required average speed of bicycle rider is 12 km/h.

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Mass of egg is 65g and volume is 60cc.The mass of salt dissolved in 100cc of water so that egg just float in water is​

Answers

Answer:

108.33 grams.

Explanation:

Given that,

Mass of a egg, m = 65 g

Volume of an egg, V = 60 cc

We need to find the mass of the salt dissolved in 100cc of water so that egg just float in water.

The density will remains constant,

[tex]\dfrac{m_1}{V_1}=\dfrac{m_2}{V_2}\\\\m_2=\dfrac{m_1V_2}{V_1}\\\\m_2=\dfrac{65\times 100}{60}\\\\=108.33\ g[/tex]

So, the required mass is 108.33 grams.

During an isothermal process one mole of a monoatomic gas did 3000 J of work on its surrounding. The final volume and pressure of the gas are 25 L and 1 atm, respectively. What was the initial volume of the gas

Answers

Answer:  The initial volume of the gas is 7.72 L

Explanation:

For an isothermal process the temperature is constant.  

[tex]PV=nRT[/tex]

as P = pressure = 1 atm ,

V = Volume =  25 L

n = moles

R= gas constant

T = temperature

[tex]PV=1atm\times 25L[/tex]

[tex]nRT=25Latm=25\times 101.3J=2532.5J[/tex]     (1Latm=101.3 J)

For isothermal reaction :

[tex]w=-2.303nRT\log\frac{V_2}{V_1}[/tex]

where , w = work done by  system = -ve

n = moles = 1

[tex]V_2[/tex] = final volume = 25 L

[tex]V_1[/tex] = initial volume = ?

[tex]-3000J=-2.303\times 2532.5\log \frac{25}{V_1}[/tex]

[tex]V_1=7.72L[/tex]

Thus initial volume of the gas is 7.72 L

A ball is thrown with a speed of 100 ft/s in a direction of 30 degrees above the horizontal. determine the horizontal distance in m, the time of flight, and the height to which it rises..Note: you have to label your answer and final answer must be two decimal place​

Answers

Answer:

Horizontal distance covered is  270.63ft

Time of flight is 3.125secs

Height to which it rises is 39.06feet

Explanation:

The horizontal distance covered is the range;

Range = U²sin2theta/g

Range = 100²sin2(30)/32

Range = 10000sin60/32

Range = 10000(0.8660)/32

Range = 8660/32

Range = 270.625ft

Hence the  horizontal distance covered is  270.63ft

The time of flight is expressed using the formula;

T = 2Usin theta/g

g is the acceleration due to gravity

T = 2(100)sin30/32

T = 200(0.5)/32

T = 100/32

T = 3.125secs

Hence the time of flight is 3.125secs

Maximum height H = u²sin²theta/2g

H = 100²(sin30)²/2(32)

H = 10000(0.5²)/64

H = 10000(0.25)/64

H = 2500/64

H = 39.0625

Hence the height to which it rises is 39.06feet

Please help

4. What are the lowest points on a transverse wave called?
a. Crests
b. troughs
C. compressions
d. rarefractions


5. Any substance that a wave moves through is called a
a. medium
b. vibrate
C. crest
d. frequency

Answers

Answer:

Explanation:

the lowest points of a transverse wave are called the troughs what is the wavelength of a wave traveling through a rope if the distance from one crest to the next is 1 meter a. 2 m

4. Low points are called troughs

5. Medium: substance through which a wave can travel

Which of these is considered a contact force?

a the force between two magnets.
b the gravitational pull of a planet.
c the friction between an object and air.
d The force between two charged particles.

Answers

Answer:

C -) The friction between an object and air.

Explanation:

The frictional force is the force that exists between two surfaces in contact, which is opposed to the movement.

That is, this type of force exists as long as there is physical contact between surfaces.

While the other types of force always act at a separation distance between bodies.

What is the magnitude of the velocity after it hits the
ground?
9.3 m/s
12 m/s
41 m/s
73 m/s

Answers

Answer: 9.3m/s

Explanation:

Your question isn't complete but let me help out:

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?

We would use Newton's law of motion to solve this which goes thus:

F = ma

f = m(v-u)/t

Cross multiply

ft = m(v-u)

where,

f = 55

t = 45/1000 = 0.045

m = 0.0060

u = -32

v = Unknown

Therefore,

55 × 45/1000 = 0.060(v - -32)

55 × 0.045 = 0.060(v + 32)

2.475 = 0.06(v + 32)

2.475 = 0.06v + 1.92

0.06v = 2.475 - 1.92

0.06v = 0.555

v = 0.555/0.06

v = 9.25m/s

v = 9.3m/s Approximately

Answer:

A.9.3 m/s

Explanation:

A rotating space station simulates artificial gravity by means of centripetal acceleration at the rim. The radius of the station is 1156 m and the apparent acceleration of gravity at the rim is 6.13 m/s2. What is the rotation rate of the station in rpm (revolutions per minute).

Answers

a= 1156× 6.13= 7086.28

Explanation:

a = m× m/ s^2

How would taping a coin to the balloon affect the overall motion of the balloon?

Answers

Answer:

the motion of the coin taping the balloon is the balloon squshing down

A person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?

Answers

Answer:

3.88 × 10^-4 m

Explanation:

Given that a  person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?

The total weight = 614 + 85

The total weight = 699N

Let the total area of contact = A

Pressure = Force / A

Substitute all the parameters into the formula

900000 = 699 /A

A = 699 / 900000

A = 7.77 × 10^-4 m

The area of contact between each tire and the ground will be = A/2

That is, 7.77 / 2 = 3.88 × 10^-4 m

Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.

diffine thermodynamics ​

Answers

Answer:

The science of the conversions between heat and other forms of energy is thermodynamics.

Hope it helps :)

Answer:

thermodynamic is the branch of physics that deals with the  relationship between heat and  other forms of energy .it describes  how thermal energy is converted to  other forms  and how it will affect matter.

Explanation:

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