Calculate the wavelength (in m) of the note middle "C" (frequency = 256 Hz) in air at room temperature and pressure.

The conscious mind contains all of the thoughts, memories, feelings, and wishes of which we are aware at any given moment. This is the aspect of our mental processing that we can think and talk about rationally. This also includes our memory, which is not always part of consciousness but can be retrieved easily and brought into awareness.

hope it helps...!!!

According to Sigmund Freud, memories that easily surface into everyday thought are a conscious part of the mind.

At any given time, the conscious mind contains all of our thoughts, memories, feelings, and wants.

This is the element of our brain functioning that we can reasonably think about and discuss. This includes our memories, which are not always conscious but can be readily recovered and brought into awareness.

Memories that easily emerge into everyday consciousness, according to Sigmund Freud, are a conscious element of the mind.

Hence, memories that easily surface into everyday thought are a conscious part of the mind.

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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:

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.

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:

Explanation:

You can use [tex]Q = mc\Delta T=(550)(4.184)(28.8-12.7)=37049.32[/tex] joule.

Based on research by Abraham Maslow, the meta-needs which are fulfilled by "making more money" are none of the meta-needs.

According to Abraham Maslow, meta-needs are human needs that relate to beauty, knowledge and creativity.

All of these can be met without having to make more money so it stands to reason that there are no meta-needs that can be met by making more money.

The main need that is met by making more money according to Maslow is Security and it is not a meta-need.

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Answer:

proton is positively charged changechar

Explanation:

Answer:

when it has a fewer amount of electrons than protons

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:

it allows electric charges to travel through freely

The pictorial diagram with a rocket launched straight upward, showing the first three positions of the rocket with equal spacing between them, best illustrates projectile motion. In this case, the rocket is projected upward into the air, and its path follows a parabolic trajectory due to the influence of gravity.

Projectile motion refers to the motion of an object that is thrown, launched or otherwise projected into the air and then moves under the influence of gravity alone. In this type of motion, the object moves along a curved path, called a parabola, due to the combined effect of its initial velocity and the force of gravity acting on it.

When a rocket is launched straight upward, it experiences a projectile motion. The rocket's initial velocity is upwards, but gravity causes it to decelerate until it eventually stops at the highest point of its trajectory. The rocket then starts to fall back down, accelerating due to gravity until it reaches the ground. If we take three equal intervals of time, we can show the rocket's positions at those times. Initially, the rocket will be moving upwards with a decreasing velocity, then it will reach its maximum height before falling back down with an increasing velocity.

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Answer: I believe it should be the final answer,

A pictorial diagram of a person hopping sideways at 5 different positions. The start and end positions are on the ground. The third image is the highest off the ground

Explanation:

Answer:

150   kg-m/s

Explanation:

momentum = m * v

50 * 3 = 150

Answer:

Explanation:

Answer: Sonic Boom.

When the object reaches the speed of sound, air cannot readily move out of the way and a shock wave is formed. When the object is moving faster than sound, the resulting sounds travel behind the object, creating a sonic boom.

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:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

Answer: This is because the force of gravity is much less on the moon than on the earth, therefore the person won't be pulled down much and will jump higher.

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:

787.5m/s

Explanation:

Speed = wavelength x frequency

speed = 2.25m x 350 Hz

= 787.5m/s

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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According to a periodic table, Krypton was created during the fission of Uranium.

Atomic number is a characteristic associated with an element and indicates its number of protons, when a fision occurs, the total number protons is conserved.

Thus, the fission of uranium is led by two elements with atomic numbers 56 and 36. According to a periodic table, those atomic numbers are associated to elements Barium (Ba) and Krypton (Kr), respectively.

According to a periodic table, Krypton was created during the fission of Uranium. [tex]\blacksquare[/tex]

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Answer:

Explanation:

The point P is influenced by two electric fields, that are from Q1 and Q2. Because Q1 is a positive point charge, the E-fields is going to the right and the E-fields from Q2 is going to the left because of negative point charge. So:

[tex]E_{1}=\frac{kQ_{1}}{R_{1}^{2}}[/tex]

[tex]E_{2}=\frac{kQ_{2}}{R_{2}^{2}}[/tex]

Where R1 = 2 + 3 = 5 m, and R2 = 3 m. With k = [tex]k=9\times 10^{9}[/tex] in [tex]Nm^{2}/C^{2}[/tex], we can obtain:

[tex]E_{1}=(9\times 10^{9})\frac{500\times10^{-6}}{25}=180\times 10^{3}[/tex] N/C (+)

[tex]E_{2}=(9\times 10^{9})\frac{100\times 10^{-6}}{9}=100\times 10^{3}[/tex] N/C (-)

So: the Electric Field at P : [tex]E_{P}=E_{1}-E_{2}=(180-100)\times 10^{3}=8\times 10^{4}[/tex] N/C in magnitude

Answer:

125J

Explanation:

[tex]work \: = force \: \times distance \\ = 25 \times 5 \\ = 125joules[/tex]

Answer:

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

Explanation:

number of protons

the atomic number of an atom is equal to the number of protons

Answer: I would buy the thermometer with the most features because they all cost the same price so why not you getting more than what you are paying for than buying something the same but with less feature and things it can do for the same price.

Hope this helps :) If so brainliest would mean alot thanks!

Explanation:

Hi there!

A)

At this point, the pilot is experiencing a DOWNWARD NET centripetal force (ALWAYS points towards the center of the circle) as well as a downward force due to gravity. Additionally, we can say that the normal force exerted by the seat on the pilot is its apparent weight.

[tex]\Sigma F = F_g + F_N[/tex]

**The net force is the centripetal force.

We can assign both to be positive since they work in the same direction. (Assigning down as positive in THIS situation will make things easier).

We know that:
[tex]F_c = \frac{mv^2}{r}[/tex]

Fc = Centripetal force (N)
m = mass (kg)
v = velocity (m/s)

r = radius (m)

[tex]F_g = mg[/tex]

Fg = Force due to gravity (N)
m = mass (kg)
g = acceleration due to gravity (9.81 m/s²)

Solve for mass. (Given weight)

[tex]m = \frac{F_g}{g}\\\\m = \frac{719}{9.8} = 73.367 kg[/tex]
Now, we have everything we need to solve.

[tex]\frac{mv^2}{r} = F_g + F_N \\\\\frac{mv^2}{r} - F_g = F_N\\\\F_N = \frac{(73.367)(225^2)}{3110} - 719 = \boxed{475.278 N}[/tex]

B)
Weightlessness is achieved when the pilot is in free fall. In this instance, FN (Normal force by the seat) = 0 N.

We can use the same setup as above.

[tex]\frac{mv^2}{r} = F_g + 0 (F_N)[/tex]

[tex]\frac{mv^2}{r} = mg[/tex]

Solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v = \sqrt{gr} = \sqrt{(9.81)(3110)} = \boxed{174.669 \frac{m}{s}}[/tex]

C)
At the BOTTOM, we have the force of gravity working DOWNWARD (Negative in this instance), while the centripetal force (NET FORCE) and normal force are upward (Positive). Do another summation of forces:
[tex]\Sigma F = F_N - F_g\\\\\\frac{mv^2}{r} + F_g = F_N \\\\F_N = \frac{(73.367)(225^2)}{3110} + 719 = \boxed{1913.278 N}[/tex]

Answer:

[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

[tex]|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb[/tex]

[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]

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If steam in a cylinder expands against a piston, exerting 10 atm of external pressure, the change in energy of the steam is mathematically given as

dE= -3010 kJ

Generally, the equation for the workdone  is mathematically given as

Work = - (pressure * Volume)

Therefore

w= - (10 x 10)

w= -100 L atm

In conclusion

dEnergy = q + work

dE = -3000 + -10.13

dE= -3010 kJ

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