✓ 4.) A certain molecule has an energy-level diagram for its vibrational energy in
which two levels are 0.0141eV apart. Find the wavelength of the emitted
line for the molecule as it falls from one of these levels to the other. [h
6.63 x 10-34 kgm s-?, C = 3 x 108ms-1, lev = 1.602 x 10^-19j

Answer:

About 5000 years.

Explanation:

Answer:

Explanation:

it will take more than half a million years for the photon to escape the sun. If you think it's about a centimeter, then it will take about 5,000 years for the photon to get outside the su

Answer:

The tone really matters and if there are any exclamation marks also.

Explanation:

Answer:

overtone- one over the first

n skips by twos

4 antinodes

500 Hz

Explanation: Hope this helps :)

Answer:

First truck has greater momnetum

Explanation:

momentum = m * v

truck a    5000 * 20 = 10 0000

truck b   1000 * 50   = 5 0000

Answer:

0.52     8.9  

Explanation:

calc.

The tension in the rope just before the student releases it is 576 N.

Tension is defined in physics as the pulling force transmitted axially by a string, rope, chain, or similar object.

Just before the student releases the rope, his velocity is zero. At this point, all of his kinetic energy has been converted into potential energy. Therefore, we can write:

1/2 m [tex]v^2[/tex] = mgh

h = [tex]v^2[/tex] / (2g) = [tex](6.2 m/s)^2[/tex] / (2 x 9.8 [tex]m/s^2[/tex]) ≈ 1.95 m

m g = 56 kg x 9.8 [tex]m/s^2[/tex] ≈ 549 N

The tension required to keep the rope taut is equal to the horizontal component of the force due to the weight of the student, which is:

T = m g / cos(θ)

tan(θ) = h / L

tan(θ) = 1.95 m / 10.0 m ≈ 0.195

θ ≈ 10.9°

So, T = m g / cos(θ) ≈ 576 N

To find the maximum tension in the rope during the swing, we can use conservation of angular momentum.

L = I ω

I = (1/3) m [tex]L^2[/tex]

I = (1/3) x 10.0 kg x [tex](10.0 m)^2[/tex] ≈ 3333.3 kg[tex]m^2[/tex]

The angular velocity of the rope just after the student releases it is:

ω = v / L

ω = 6.2 m/s / 10.0 m ≈ 0.62 rad/s

L = I ω ≈ 2070 N m s

Thus, the angular momentum of the system just after the student releases the rope is 2070 N m s.

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Answer:

Mass.

Explanation:

I took the quiz and got the answer right

Answer:

B: The object will start moving

Explanation:

If energy is transferred the object will definitely change so it can't be a. If you add energy the object will have more force so it cant be c. The mass of an object can't increase just by giving an object energy so it cant be d

200 Joules should be supplied in the circuit.

Here the output is 30J and input is missing

[tex]\sf \dfrac{30}{I} * 100 = 15[/tex]

multiplied with 100, to covert it into percentage

[tex]\sf 3000 = 15(I)[/tex]

[tex]\sf I = \dfrac{3000 }{15}[/tex]

[tex]\sf I = 200J[/tex]

Let that be X

[tex]\\ \rm\Rrightarrow \dfrac{30}{x}(100)=15[/tex]

[tex]\\ \rm\Rrightarrow 3000=15x[/tex]

[tex]\\ \rm\Rrightarrow x=200J[/tex]

Answer:

The circuits that were made previously were large and bulky, consisting of circuit components like resistor, capacitor, inductor, transistor, diodes, etc., which were connected with copper wires.

Explanation:

I am in confusion with mono electronic circuit and I am not want to made u confused person.

I am really sorry brother.

:-befrank

Explanation:

Since the acceleration is constant, we use the big four equations.

[tex]x = x _{i} + vt[/tex]

[tex]x = \frac{1}{2} a {t}^{2} + v _{i}t + x _{i}[/tex]

[tex]v {}^{2} = (v _{i}) {}^{2} + 2a(x - x _{i})[/tex]

[tex] \frac{1}{2} (v + v _{i})t = x - x _{i}[/tex]

Next. identity the variables.

We know

t=9.2

change in position or (x- x_i) is 428.8

Initial Velocity is 0.

We need to solve for the final velocity, so we use the fourth equation.

[tex] \frac{1}{2} (v - v _{i})t = x - x _{i}[/tex]

Subsitue

[tex] \frac{1}{2} (v - 0)(9.2) = 428.8[/tex]

[tex] \frac{1}{2} (v)9.2 = 428.8[/tex]

[tex]v = 93.22 \frac{m}{s} [/tex]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics.

Answer:

When a body is immersed fully or partially in a fluid, it experience is an upward force that is equal to the weight of the fluid displaced by it.

Answer:

2.25 x 10^-6   ohm - m     2.25 x 10^-4 ohm - cm

Explanation:

rho = R A/l

  R = .040

   A = .015 * .015  m^2

     l = 4         rho =

Answer:

Find the heavy processes of CO2 and methane.

Explanation:

Step #1 Find a very reliable energy source.

Step #2 Find the heavy processes of CO2 and methane like in many factories and places with high carbon emissions.

Step #3 Shut down each of the heavy processes even 10 per day would do a lot.

Step #4 Use the renewable energy source as a replacement for the attempts to generate energy using carbon emissions.

Answer:

No

Explanation:

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

Answer:

150   kg-m/s

Explanation:

momentum = m * v

50 * 3 = 150

Answer:

proton is positively charged changechar

Explanation:

Answer: The maximum kinetic energy KEe of ejected electrons (photoelectrons) is given by KEe = hf − BE, where hf is the photon energy and BE is the binding energy (or work function) of the electron to the particular material.

Explanation:

The work function of the uranium metal will be given by the value 7.9*10^-19

The work function of a particle is defined as the product of the planks constant to the frequency of the particle.

[tex]\phi=h\times f[/tex]

Now the frequency will be given as

[tex]f=\dfrac{c}{\lambda}[/tex]

So the formula will become

[tex]\phi=\dfrac{hc}{\lambda}[/tex]

h=6.626 × 10-34 J∙s

c=3.0 × 108 m/s

[tex]\lambda[/tex] = 250 nm 250* 10^-9

[tex]\phi=\dfrac{6.626\times 10^{-26}\times3\times 10^{8}}{250\times 10^{-9}}[/tex]

[tex]\phi=7.9\times 10^{-19}[/tex]

Hence the work function of the uranium metal will be given by the value 7.9*10^-19

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Answer:

Explanation:

Distributive property:

A(B+C)=AB+AC

(4x-2)2

Switch sides.

→ 2(4x-2)

Multiply.

⇒ 2*4x=8x

⇒ 2*2=4

Rewrite the problem down.

⇒ = 8x-4

[tex]\Longrightarrow: \boxed{\sf{8x-4}}[/tex]

I hope this helps you! Let me know if my answer is wrong or not.

Answer: 8x - 4

Explanation:

(4x -2)*2

apply distributive property

2* 4x - 2 * 2

Multiply the monomials = 8x

2*4x - 2*= 8x-2*2

calculate =

8x-4

Answer: 8x - 4

Answer:

20 or 12

Explanation:

if it is 20 u will have to multiple 6 by 4 and the answer u will multiple it by2 and when it is 12 u will have to add 6 +4 and the answer u will add it to 2

Answer:

The circular turning of roads

Explanation:

please mark brainliest

Answer:

Explanation:

The circular turning of roads

Driving on Curves

Answer:

A. length, diameter, material, temperature

Answer:

[tex] \red{ \rule{1pt}{500000pt}}[/tex] [tex] \red{ \rule{1pt}{500000pt}}[/tex]

Explanation:

Waves that make up the radiation collide with each other so that they add together or cancel each other out.

This is the phenomenon which occurs when two waves meet while traveling along the same medium.

In the double-slit test experiment, bright and dark fringes was observed which was caused by the superposition of overlapping light waves originating from the two slits which is why option A was chosen as the most appropriate choice.

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nothing

Explanation:

we can say that it was certainly bad for the shopkeeper obviously and we should not be making questions about the physics behind that accident and should call the cops or 911

a. The net force from t = 1 to t = 6 s is 112 N

b. The net force from t = 7 to t = 12 s is 0 N

Force is that thing which causes a change in motion of an object. The net force is given mathematically as F = ma where

This is the resultant sum of all the forces acting on an object. It is also given by F = ma where

The net force from t = 1 to t = 6 s is 112 N

From the table from time t = 1 to t = 6 s, the acceleration a = 1.4 m/s². Since m = 80 kg,

F = ma

= 80 kg × 1.4 m/s²

= 112 kgm/s²

= 112 N

So, the net force from t = 1 to t = 6 s is 112 N

The net force from t = 7 to t = 12 s is 0 N

From the table from t = 7 to t = 12 s, the acceleration a = 0 m/s²

Since m = 80 kg,

F = ma

= 80 kg × 0 m/s²

= 0 kgm/s²

= 0 N

So, the net force from t = 7 to t = 12 s is 0 N

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Answer:

See below

Explanation:

-4 rev / min^2 * 55 min =  - 220 rev/ min

480 - 220 = 260 rev/min at end

2pi radians /rev * 260 rev/min  = 1633.6 Radians/min

At the end of the music, the angular speed of the C.D. would be (d) 260 rev/minute and 1633.6 rad/minute

To calculate angular speed following information is given in the question:

Hence, option (d) is correct which states that the angular speed is 260 rev/min and 1633.6 radians/min.

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Answer:

Chapter 1

1. Show that the Lorentz transformation is such that the velocity of a light ray

travelling in the x direction is the same for the observer in the frame S and for

the observer in the frame S

.

Solution: Consider a light ray travelling in the x direction. If the light ray

connects two space–time points {t1, x1} and {t2, x2}, we have

c = x2 − x1

t2 − t1

The speed of light observed in the frame S will be

c = x

2 − x

1

t

2 − t

1

= c

γ ((x2 − x1) − βc(t2 − t1))

γ (c(t2 − t1) − β(x2 − x1))

= c

x2 − x1

t2 − t2

− βc

c − β x2 − x1

t2 − t2

= c

2. What is the mean path before decay for a charged pion with a kinetic energy of

1 GeV?

Solution: The pion has a lifetime 2.6 × 10–8 s and a mass of 139.6 MeV. If the

energy is 1 GeV, the velocity of the pion is 99% of the velocity of light (Eq. 1.4).

The mean path before decay is

= 0.99 c γ τ

= 0.99 c

1000 + 139.6

139.6

2.6 10−8 = 63 m

S. Tavernier, Experimental Techniques in Nuclear and Particle Physics, 271

DOI 10.1007/978-3-642-00829-0, C Springer-Verlag Berlin Heidelberg 2010

272 Solutions to Exercises

3. Show that the relativistic expression for the kinetic energy of a particle (Eq. 1.2)

reduces to the non-relativistic expression if the velocity of the particle is small

compared to the velocity of light.

Solution:

E = Ekinetic + m0c2 = m0c2

1 − (v/c)2

≈ m0c2

(1 − 1/2(v/c)2) ≈ m0c2(1 + 1/2(v/c)

2)

= m0c2 +

1

2

m0v2

4. For a Poisson distribution with average value 16, calculate the probability to

observe 12, 16 and 20 as measured value. Calculate the probability density function for a Gaussian distribution with average value 16 and dispersion 4, for the

values x = 12, 16 and 20. Compare the results.

Solution: For a Poisson distribution P(12) = 0.0829, P(16) = 0.1024, P(20) =

0.0418

For a Gaussian distribution, f(12) = 0.0605, f(16) = 0.0997, f(20) = 0.0605

5. Consider a very short-lived particle of mass M decaying into two long-lived particles 1 and 2. Assume you can measure accurately the energies and momenta of

the two long-lived particles. How will you calculate the mass of the short-lived

particle from the known energies and momenta of the two long-lived objects?

Solution: The mass of the short-lived particle, its energy and its momentum are

related by Eq. (1.1). The energy and momentum of the particle are equal to the

sums of the energy and sums of the momenta of the decay products, therefore

M2c4 = (E1 + E2)

2 − c2(P1 + P2)

2

6. Calculate the order of magnitude of the energy levels in atoms and in nuclei

using the ‘particle in a box’ approximation, Eq. (1.9). Use for the dimension of

the atom 10–10 m and for the dimension of the nucleus 10−15 m.

Solution: Atomic energy levels: ≈40 eV; nuclear energy levels: ≈400 MeV.

7 . Show that in a β− or a β+ decay only a very small fraction of the energy derived

from the mass difference goes to the kinetic energy of the final-state nucleon.

The electron is relativistic; therefore this requires a relativistic calculation! Hint:

the 3-body problem can be reduced to a 2-body problem by considering the

electron–neutrino system as one object with a mass of a few MeV.

Solution. Consider the 2-body decay of some heavy object with mass M into

two objects with masses m1 and m2. The kinetic energy of each of the final-state

particles in the overall centre of mass system is found as follows.

Solutions to Exercises 273

Consider two particles with energy and momentum four vectors p1 and p2.

The symbol pi stands for the four-vector {Ei,cpi}. The energy E appearing in this

expression is the total energy E, i.e. the rest energy mc2 plus the kinetic energy.

The four-vector product (p1.p2) is defined as

(p1.p2) =

(

E1E2 − c2 p1 p2

)

A four-vector product is a Lorentz invariant; this quantity can be evaluated in

any reference frame, and the result is the same. Consider now the quantity

(p1.p2)

m1c2

This is a Lorentz invariant. Evaluating this expression in the rest frame of

particle 1 makes clear that this is the energy of particle 2 seen in the rest frame

of particle 1. This remains true also if one of the particles is in fact a system

of particles, for example the system of the two particles 1 and 2. The energy of

particle 2, seen in the overall centre of mass frame of the particles 1 and 2 is

therefore

E∗

2 = (p1 + p2).p2

(p1 + p2)2

We have the following relations:

(p1 + p2)

2 = M2c4

(p1.p2) = 1

2

(

(p1 + p2)

2 − (p1)

2 − (p2)

2

)

= M2c4 − m2

1c4 − m2

2c4

And therefore finally

E∗

2 = M2c4 + m2

2c4 − m2

1c4

2Mc2

Let us now apply the above relation to the decay

N∗ → N + e− + ¯νe + Q

The symbol Q represents the energy liberated in the reaction. Let us denote

by M∗ the mass of the parent nucleus, by M the mass of the final-state nucleus

and by m the mass of the electron–neutrino system. The kinetic energy of the

nucleus in the final state is given by

274 Solutions to Exercises

Ekin = M∗2c4 + M2c4 − m2c4

2M∗c2 − Mc2

= M∗2c4 + M2c4 − m2c4 − 2M∗c2Mc2

2M∗c2

= (M∗ − M)

2 c4 − m2c4

2M∗c2

=

!

mc2 + Q

The energy of the cosmic ray photon is zero. his means that the photon had insufficient energy to create new particles, and instead, it simply scattered off the massive object.

Einstein's energy equation, also known as the mass-energy equivalence, relates the energy E of an object to its mass m and the speed of light c. The equation is:

E = mc^2

where:

E is the energy of the object in joules (J)

m is the mass of the object in kilograms (kg)

c is the speed of light in meters per second (m/s)

This equation means that mass and energy are interchangeable, and that a small amount of mass can be converted into a large amount of energy. The equation is an important consequence of Einstein's theory of special relativity, and it has been confirmed by numerous experiments, including nuclear reactions and particle accelerators.

Here in the Question,

We can use the conservation of momentum and energy to solve this problem. Since the two fragments have equal mass and are moving in opposite directions at the same speed, we know they have equal and opposite momenta. Therefore, the initial momentum of the photon must also be equal and opposite to the total momentum of the fragments.

Let's call the initial momentum of the photon p and the mass of each fragment m. The total momentum of the fragments is:

p' = 2mv

where v is the speed of each fragment, which we know is 0.6c. Therefore, we can write:

p' = 2m(0.6c) = 1.2mc

By conservation of momentum, we have:

p = -p'

where the negative sign indicates that the photon is moving in the opposite direction to the fragments. Therefore:

p = -1.2mc

Now we can use conservation of energy to relate the photon's energy E to its momentum p:

E^2 = p^2c^2 + m^2c^4

Substituting the expression we found for p, we get:

E^2 = (1.2mc)^2c^2 + m^2c^4

E^2 = 1.44m^2c^4 + m^2c^4

E^2 = 1.45m^2c^4

Solving for E, we get:

E = mc^2 * sqrt(1.45)

Plugging in the values for m and c, we get:

E = (0 * 9.0 × 10^16 kg) * sqrt(1.45) = 0

Therefore, The photon from a cosmic ray has no energy. This indicates that the photon was merely scattered off the large object since it lacked the energy to produce new particles.

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​

Answer:

37500 N

Explanation:

F = ma

F = 7500 (5 m/s^2) = 37500 N