Use Snell's Law to solve the following:

Answer:

1, 2, 4, 6

Explanation:

Move the magnet forward.

Move the loop away from the magnet.

Move the loop sideways.

Rotate the loop in place.

Answer:

Explanation:

When a changing magnetic flux linked with the coil, then the induced emf or induced current is developed.

1. Move the magnet forward.

Here, the flux linked with the coil changes, so the induced current is developed.

2. Move the loop away from the magnet.

Here, the flux linked with the coil changes, so the induced current is developed.

3. Place the loop right next to the magnet, both stationary.

Here, the flux linked with the coil does not change, so the induced current is not developed.

4. Move the loop sideways.

Here, the flux linked with the coil changes, so the induced current is developed.

5. Move the magnet and loop in the same direction, at the same speed, at the same time.

Here, the flux linked with the coil does not change, so the induced current is not developed.

6. Rotate the loop in place.

Here, the flux linked with the coil changes, so the induced current is developed.

Answer:

[tex]3.8\cdot 10^{-16}N[/tex]

Explanation:

For a charged particle moving perpendicularly to a magnetic field, the magnitude of the force exerted on the particle is:

[tex]F=qvB[/tex]

where

q is the magnitude of the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

In this problem, we have

[tex]q=e=1.6\cdot 10^{-19}[/tex] is the charge of the electron

[tex]v=6.0\cdot 10^6 m/s[/tex] is the velocity

[tex]B=4.0\cdot 10^{-4}T[/tex] is the magnetic field

Substituting into the formula, we find the force:

[tex]F=(1.6\cdot 10^{-19} C)(6.0\cdot 10^6 m/s)(4.0\cdot 10^{-4}T)=3.8\cdot 10^{-16}N[/tex]

The force exerted on the electron by the magnetic field, calculated using F = qvB sin θ, is 3.84 x 10^-16 N.

The magnitude of the force exerted on a charged particle moving in a magnetic field can be calculated using the formula F = qvB sin θ, where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and magnetic field vectors. In this case with an electron moving in a straight path within a magnetic field, the angle is 90 degrees, so sin θ equals 1. The charge of an electron (q) is 1.6 x 10-19 C (Coulombs).

Therefore, substituting these values into the equation gives: F = (1.6 x 10-19 C) * (6.0 x 106 m/s) * (4.0 x 10-4 T) * 1 = 3.84 x 10-16 N (Newtons).

This means that the magnitude of the force exerted on the electron by the magnetic field is 3.84 x 10-16 N.

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Answer:

[tex]9.9\cdot 10^{-9}m[/tex]

Explanation:

In a single-slit diffraction pattern, the location of the first minimum is given by

[tex]y=\frac{L\lambda }{a}[/tex]

where

L is the distance between the slit and the screen

[tex]\lambda[/tex] is the wavelength

a is the width of the slit

In this problem, we have

[tex]y=0.527 cm = 5.27\cdot 10^{-3} m[/tex] (location of first minimum)

L = 1.068 m (distance of the screen)

[tex]a=2.00\mu m=2.00\cdot 10^{-6}m[/tex] (width of the slit)

Solving the equation for [tex]\lambda[/tex], we find the De Broglie wavelength of the electron:

[tex]\lambda = \frac{ya}{L}=\frac{(5.27\cdot 10^{-3} m)(2.00\cdot 10^{-6} m)}{1.068 m}=9.9\cdot 10^{-9}m[/tex]

The question examines the physics concept of diffraction about a beam of electrons passing through a narrow slit. De Broglie's hypothesis applies in this context as particles can exhibit characteristics of waves. The wavelength of the electron wave can be determined using the formula for the location of first-intensity minima, taking into consideration the values provided.

The question refers to a phenomenon known as diffraction, which is observed when electrons pass through a narrow slit toward a detector and create a diffraction pattern. The principle in question is the de Broglie hypothesis which maintains that particles can exhibit wave-like behavior. To calculate the wavelength λ of one of the electrons in this beam, we use the location of the first intensity minima formula: y=Lλ/a. In the given scenario, y is the distance from the center of the intensity minima to the broad maximum (0.527 cm), L is the distance to the screen (1.068 m), and a is the width of the slit (2.00μm). Using these values, you can solve for λ, the wavelength of the electron wave.

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Statement C about electric and magnetic fields is true. Option C is correct.

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained, it is the field felt around a moving electric charge.

There are two poles of the magnet;

1. North Pole.

2. South Pole.

It has two poles, one north, and one south, and when hanging freely, the magnet aligns itself such that the north pole faces the earth's magnetic north pole.

They point away from the north pole and towards the south pole, is the statement correctly describing magnetic field lines around a magnet.

Statement about electric and magnetic fields is true is;

Magnetic fields must start at the north pole of a magnet and end at the south pole of a magnet.

Hence, option C is correct.

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Answer:

[tex]1.34\cdot 10^{-16} C[/tex]

Explanation:

The strength of the electric field produced by a charge Q is given by

[tex]E=k\frac{Q}{r^2}[/tex]

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

[tex]E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C[/tex]

and the fish can detect the electric field at a distance of

[tex]r=63.5 cm = 0.635 m[/tex]

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

[tex]Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C[/tex]

Final answer:

A point charge of approximately 1.35 × 10-13 coulombs would be required to produce an electric field change of 3.00 µN/C at a distance of 63.5 cm from the elephant nose fish.

Explanation:

To calculate the amount of charge needed to produce a change in the electric field of 3.00 µN/C at a distance of 63.5 cm, we can use Coulomb's law, which describes the electric field E created by a point charge Q. Coulomb's law is given by the equation E = kQ/r2, where k is Coulomb's constant (8.987 × 109 N m2/C2), Q is the charge in coulombs, and r is the distance from the charge in meters. To find Q, we rearrange the equation to Q = Er2/k.

By substituting the values, E = 3.00 µN/C (or 3.00 × 10-6 N/C), and the distance r = 63.5 cm (or 0.635 m), we can calculate the charge Q necessary to produce the observed electric field at the specified distance.

Calculation:

Q = (3.00 × 10-6 N/C) × (0.635 m)2 / (8.987 × 109 N m2/C2)

  = 1.35 × 10-13 C

So, a point charge of approximately 1.35 × 10-13 coulombs would be needed to produce an electric field change of 3.00 µN/C at a distance of 63.5 cm from the elephant nose fish.

1. [tex]7.95\cdot 10^6 J[/tex]

The total energy given to the cells during one pulse is given by:

[tex]E=Pt[/tex]

where

P is the average power of the pulse

t is the duration of the pulse

In this problem,

[tex]P=1.59\cdot 10^{12}W[/tex]

[tex]t=5.0 ns = 5.0\cdot 10^{-9} s[/tex]

Substituting,

[tex]E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J[/tex]

2. [tex]1.26\cdot 10^{21}W/m^2[/tex]

The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is

[tex]r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m[/tex]

So the area of each cell is

[tex]A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2[/tex]

The energy is spread over 100 cells, so the total area of the cells is

[tex]A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2[/tex]

And so the intensity delivered is

[tex]I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2[/tex]

3. [tex]9.74\cdot 10^{11} V/m[/tex]

The average intensity of an electromagnetic wave is related to the maximum value of the electric field by

[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

E is the amplitude of the electric field

Solving the formula for E, we find:

[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m[/tex]

4. 3247 T

The magnetic field amplitude is related to the electric field amplitude by

[tex]E=cB[/tex]

where

E is the electric field amplitude

c is the speed of light

B is the magnetic field

Solving the equation for B and substituting the value of E that we found at point 3, we find

[tex]B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T[/tex]

Answer:

(B)

Explanation:

While some of the other scenarios make sense to match with Newton's Third Law of Motion, their reaction and actions are in the wrong order.

The scenario: An ax being used to hit wood (action) and the wood splitting after being hit by the ax (reaction), accurately names the correct action and reaction pair to apply to Newton's Third Law.

To find the correct option among all the options, we need to know about the Newton's third law.

Newton's third law of motion says that every action has a opposite and equal reaction.

Thus, we can conclude that the option (b) is correct.

Learn more about the Newton's third law here:

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(a) 0.028 rad

The angular separation of the nth-maximum from the central maximum in a diffraction from two slits is given by

[tex]d sin \theta = n \lambda[/tex]

where

d is the distance between the two slits

[tex]\theta[/tex] is the angular separation

n is the order of the maximum

[tex]\lambda[/tex] is the wavelength

In this problem,

[tex]\lambda=700 nm=7\cdot 10^{-7} m[/tex]

[tex]d=0.025 mm=2.5\cdot 10^{-5} m[/tex]

The maximum adjacent to the central maximum is the one with n=1, so substituting into the formula we find

[tex]sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{2.5\cdot 10^{-5} m}=0.028[/tex]

So the angular separation in radians is

[tex]\theta= sin^{-1} (0.028) = 0.028 rad[/tex]

(b) 0.028 m

The screen is located 1 m from the slits:

D = 1 m

The distance of the screen from the slits, D, and the separation between the two adjacent maxima on the screen (let's call it y) form a right triangle, so we can write the following relationship:

[tex]\frac{y}{D}=tan \theta[/tex]

And so we can find y:

[tex]y=D tan \theta = (1 m) tan (0.028 rad)=0.028 m[/tex]

(c) [tex]2.8\cdot 10^{-4} rad[/tex]

In this case, we can apply again the formula used in part a), but this time the separation between the slits is

[tex]d=2.5 mm = 0.0025 m[/tex]

so we find

[tex]sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{0.0025 m}=2.8\cdot 10^{-4}[/tex]

And so we find

[tex]\theta= sin^{-1} (2.8\cdot 10^{-4}) = 2.8\cdot 10^{-4} rad[/tex]

(d) [tex]7.0\cdot 10^{-6} m = 7.0 \mu m[/tex]

This part can be solved exactly as part b), but this time the distance of the screen from the slits is

[tex]D=25 mm=0.025 m[/tex]

So we find

[tex]y=D tan \theta = (0.025 m) tan (2.8\cdot 10^{-4} rad)=7.0\cdot 10^{-6} m = 7.0 \mu m[/tex]

(e) The maxima will be shifted, but the separation would remain the same

In this situation, the waves emitted by one of the slits are shifted by [tex]\pi[/tex] (which corresponds to half a cycle, so half wavelength) with respect to the waves emitted by the other slit.

This means that the points where previously there was constructive interference (the maxima on the screen) will now be points of destructive interference (dark fringes); on the contrary, the points where there was destructive interference before (dark fringes) will now be points of maxima (bright fringes). Therefore, all the maxima will be shifted.

However, the separation between two adjacent maxima will not change. In fact, tall the maxima will change location exactly by the same amount; therefore, their relative distance will remain the same.

A white ring buoy appears blue because the blue plastic absorbs all colors of light except blue. Only the blue light reflected from the ring buoy passes through the blue plastic.

Answer:

blue

absorbs

reflected from

Explanation:

I looked it up and everybody said this :)

Answer: Mars

Explanation:

Mars is a planet similar to the earth which the rest can't sustain life either because it's too hot or cold or too much gas maybe even toxic acid.

Final answer:

Mars is the most likely planet other than Earth to have life in our solar system, based on past habitable conditions. The neighboring terrestrial planets, Venus and Mars, have evolved differently, with Earth being the most habitable. Gas giants like Jupiter and Saturn are not considered habitable.

Explanation:

Mars is currently thought to be the most likely planet, after Earth, to have life. Despite being colder and drier than Earth, Mars has shown evidence of habitable conditions in the past, making it a prime candidate for potential life forms.

Venus and Mars are the neighboring terrestrial planets diverged significantly in their evolution, with Earth being the most hospitable planet in the solar system. The outer gas giant planets like Jupiter and Saturn are almost certainly not habitable for life as we know it.

17.

There are three different methods for charging objects:

- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.

- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.

- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.

18.

Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)

[tex]E=k\frac{q}{r^2}[/tex]

where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.

Also, the direction of the field is determined by the sign of the charge:

- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)

- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)

19.

Electrical force is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where k is the Coulomb's constant, q1 and q2 are the two charges, and r their separation.

Gravitational force is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.

Similarities between the two forces:

- Both are inversely proportional to the square of the distance between the two objects, r

- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)

- Both forces have infinite range

Differencies between the two forces:

- The electric force can be either attractive or repulsive, while the gravitational force is attractive only

- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G

(a) 646.9 Hz

The formula for the Doppler effect is:

[tex]f'=\frac{v+v_o}{v-v_s}f[/tex]

where

f = 600 Hz is the real frequency of the sound

f' is the apparent frequency

v = 345 m/s is the speed of sound

[tex]v_o = 0[/tex] is the velocity of the observer (zero since it is stationary at the station)

[tex]v_s = +25 m/s[/tex] is the velocity of the source (the train), moving toward the observer

Substituting into the formula,

[tex]f'=\frac{345 m/s+0}{345 m/s-25 m/s}(600 Hz)=646.9 Hz[/tex]

(b) 20.1 m/s

In this case, we have

f = 600 Hz is the real frequency

f' = 567 Hz is the apparent frequency

Assuming the observer is still at rest,

[tex]v_o = 0[/tex]

so we can re-arrange the Doppler formula to find [tex]v_s[/tex], the new velocity of the train:

[tex]f'=\frac{v}{v-v_s}f\\\frac{f}{f'}=\frac{v-v_s}{v}\\\frac{f}{f'}v=v-v_s\\v_s = (1-\frac{f}{f'})v=(1-\frac{600 Hz}{567 Hz})(345 m/s)=-20.1 m/s[/tex]

and the negative sign means the train is moving away from the observer at the station.

The observed frequency of the train's horn is 600 Hz for a stationary observer. As the train moves away, its speed is calculated to be approximately 14.81 m/s to yield an observed frequency of 567 Hz.

This question revolves around the principle of Doppler Effect in Physics. The Doppler Effect explains how observed frequency shifts depending on the relative motion of the source (in this case, a train) and the observer (a person at the station).

(a) As the train approaches the person, the frequency detected will be higher than the original. This can be calculated using the formula: f' = f((v + vo) / v), where f' is the observed frequency, f is the source frequency, v is the speed of sound, and vo is the observer's velocity. Since the observer is stationary, vo = 0. f' = 600 Hz * ((345 m/s + 0) / 345 m/s) = 600 Hz

(b) As the train moves away from the person, the observed frequency decreases. Use the formula: f' = f * (v / (v + vs)), where vs is the source's velocity. Solving for vs, we get vs = v - (v * f / f'), equals to 345 m/s - (345 m/s * 600 Hz / 567 Hz) = 14.81 m/s.

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Answer:

C. Farmers and gardeners should use other methods to provide crops with the nitrogen they need.

Explanation:

That was the right answer on study island :/

Answer:

[tex]-1.14 \cdot 10^{-3} V[/tex]

Explanation:

The induced emf in the loop is given by Faraday's Newmann Lenz law:

[tex]\epsilon = - \frac{d \Phi}{dt}[/tex] (1)

where

[tex]d\Phi[/tex] is the variation of magnetic flux

[tex]dt[/tex] is the variation of time

The magnetic flux through the coil is given by

[tex]\Phi = NBA cos \theta[/tex] (2)

where

N = 6 is the number of loops

A is the area of each loop

B is the magnetic field strength

[tex]\theta =15^{\circ}[/tex] is the angle between the direction of the magnetic field and the normal to the area of the coil

Since the radius of each loop is r = 5.00 cm = 0.05 m, the area is

[tex]A=\pi r^2 = \pi (0.05 m)^2=0.0079 m^2[/tex]

Substituting (2) into (1), we find

[tex]\epsilon = - \frac{d (NBA cos \theta)}{dt}= -(NAcos \theta) \frac{dB}{dt}[/tex]

where

[tex]\frac{dB}{dt}=0.0250 T/s[/tex] is the rate of variation of the magnetic field

Substituting numbers into the last formula, we find

[tex]\epsilon = -(6)(0.0079 m^2)(cos 15^{\circ})(0.0250 T/s)=-1.14 \cdot 10^{-3} V[/tex]

Answer:

Induced emf, [tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]

Explanation:

It is given that,

Number of circular loop, N = 6

A uniform magnetic field perpendicular to its surface is held stationary at an angle of 15 degrees to the ground.

Radius of the loop, r = 5 cm = 0.05 m

Change in magnetic field, [tex]\dfrac{dB}{dt}=0.025\ T/s[/tex]

Due to the change in magnetic field, an emf will be induced. Let E is the induced emf in the coil. it is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=\dfrac{d(NBA\ cos\theta)}{dt}[/tex]

[tex]\epsilon=-NA\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=6\times \pi (0.05)^2\times 0.025\times cos(15)[/tex]

[tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]

So, the induced emf in the loop is [tex]1.13\times 10^{-3}\ volts[/tex] . Hence, this is the required solution.

1. Helium nucleus and electron/positron

- An [tex]\alpha[/tex] decay is a decay in which an [tex]\alpha[/tex] particle is produced.

An [tex]\alpha[/tex] particle consists of 2 protons and 2 neutrons: therefore, in an alpha-decay, the nucleus loses 2 units of atomic number (number of protons) and 2 units of mass number (sum of protons+neutrons).

The alpha-particle consists of 2 protons of 2 neutrons: so it corresponds to a nucleus of helium, which consists exactly of 2 protons and 2 neutrons.

- There are two types of [tex]\beta[/tex] decay:

-- In the [tex]\beta^-[/tex] decay, a neutron decays into a proton emitting a fast-moving electron and an anti-neutrino:

[tex]n \rightarrow p + e^- + \bar{\nu}[/tex]

and the [tex]\beta[/tex] particle in this case is the electron

--  In the [tex]\beta^+[/tex] decay, a proton decays into a neutron, emitting a fast-moving positron and a neutrino:

[tex]p \rightarrow n + e^+ + \nu[/tex]

and the [tex]\beta[/tex] particle in this case is the positron.

2) Gamma ray

A [tex]\gamma[/tex] decay occurs when an unstable (excited state) nucleus decays into a more stable state. In this case, there are no changes in the structure of the nucleus, but energy is released in the form of a photon:

[tex]X^* \rightarrow X + \gamma[/tex]

where the wavelength of this photon usually falls in the part of the electromagnetic spectrum corresponding to the gamma ray region.

So, the [tex]\gamma[/tex] ray is commonly known as gamma radiation.

3)

The sign of the three forms of radiation are the following:

- [tex]\alpha[/tex] particle: it consists of 2 protons (each of them carrying a positive charge of +e) and 2 neutrons (uncharged), so the total charge is

Q = +e +e = +2e

- [tex]\beta[/tex] particle: in case of [tex]\beta^-[/tex] radiation, the particle is an electron, so it carries a charge of

Q = -e

in case of [tex]\beta^+[/tex] radiation, the particle is a positron, so it carries a charge of

Q = +e

- [tex]\gamma[/tex] radiation: the [tex]\gamma[/tex] radiation consists of a photon, and the photon has no charge, so the charge in this case is

Q = 0

Answer:

98,000J

Explanation:

Given parameters :

Mass = 2000kg

Height = 5m

Time = 10s

P. E = mgh

P. E = 2000x 5x 9.8

P.E = 98,000J

The potential energy of a 2000 kg mass raised to a height of 5.0 m is calculated using the formula PE = mgh, resulting in a potential energy of 98000 Joules.

To determine the potential energy of a mass at a certain height, we can use the formula for gravitational potential energy, which is PE = mgh, where:

In this case, the mass m is 2000 kg, g is 9.8 m/s2, and the height h is 5.0 m. So the potential energy can be calculated as follows:

PE = (2000 kg)
(9.8 m/s2)
(5.0 m) = 98000 kg
m2 s−2 = 98000 Joules

Therefore, the potential energy of the mass at 5.0 m height is 98000 J.

Answer:

7 electrons

Explanation:

We can solve the problem by using the law of conservation of electric charge: in fact, the total electric charge before and after the collision must be conserved.

Before the collision, we have:

- A nucleus of carbon-12, consisting of 6 protons (charge +1 each) + 6 neutrons (charge 0 each), so total charge of +6

- A nucleus of nitrogen-14, consisting of 7 protons (charge +1 each) + 7 neutrons (charge 0 each), so total charge of +7

So the total charge before the collision is +6+7=+13 (1)

After the collision, we have:

- 17 protons (charge +1 each): total charge of +17

- 4 antiprotons (charge -1 each): total charge of -4

- 7 positrons (charge +1 each): total charge of +7

- 25 neutral particles (charge 0 each): total charge of 0

- N electrons (charge -1 each): total charge of -N

So the total charge after the collision is +17-4+7+0-N=+20-N (2)

Since the charge must be conserved, we have (1) = (2):

+13 = +20 - N

Solving for N,

N = 20 - 13 = 7

So, there are 7 electrons.

The number of electrons emitted in the collision between carbon-12 and nitrogen-14 nuclei detected as 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles, can be calculated to be 20 to maintain charge neutrality.

When the graduate student bombards nitrogen-14 nuclei with carbon-12 nuclei and detects the disintegration products including 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles, the amount of emitted electrons can be calculated by considering the charge balance of the particles. Since protons have a +1 charge and electrons have a -1 charge, positrons also have a +1 charge and antiprotons have a -1 charge, the net charge must remain the same as before the collision.

Originally, the nitrogen-14 nucleus (with 7 protons) and the carbon-12 nucleus (with 6 protons) totaled 13 positive charges. After the collision, the detected 17 protons and 7 positrons contribute +24 to the net charge, while 4 antiprotons contribute -4, making the total positive charge +20. To counterbalance this and maintain a neutral charge, 7 additional electrons (on top of the original 13) must have been emitted to bring the net charge back down to +13. Thus, in total, there must be 20 electrons emitted.

The strength of the magnetic force on your head while driving west at the equator is 4.05 x 10^{-5} N, and the direction of this force is upwards, as determined by the right-hand rule.

The force on a charge moving in a magnetic field can be calculated using the formula F = qvB, where F is the force in newtons (N), q is the charge in coulombs (C), v is the velocity in meters per second (m/s), and B is the magnetic field strength in teslas (T). Given the charge on your head is 9 x 10^{-9} C, the Earth's magnetic field strength at the equator is 5 x 10^{-5} T, and your velocity driving west is 90 m/s, the magnitude of the magnetic force can be calculated as follows: F = 9  imes 10^{-9} C x 90 m/s x 5 x 10^{-5} T = 4.05 x 10^{-5} N.

Using the right-hand rule to determine the direction of the force: if you point your thumb in the direction of the velocity (west), and your fingers in the direction of the Earth's magnetic field (north), the force (perpendicular to the palm) will point upwards. Therefore, the direction of the magnetic force on your head is upwards.

(a) 80 N/m

The spring constant can be found by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

[tex]k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m[/tex]

(b) 0.84 Hz

The frequency of oscillation of the system is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

[tex]f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz[/tex]

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

[tex]v=\omega A[/tex]

where

[tex]\omega[/tex] is the angular frequency

A is the amplitude of the motion

For this system, we have

[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]

[tex]A=0.200 m[/tex] (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

[tex]v=(5.25 rad/s)(0.200 m)=1.05 m/s[/tex]

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex]

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

[tex]a=\omega^2 A[/tex]

Using the numbers we calculated in part c):

[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]

[tex]A=0.200 m[/tex]

we find immediately the maximum acceleration:

[tex]a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2[/tex]

(f) At the position of maximum displacement: [tex]x=\pm 0.200 m[/tex]

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

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

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

[tex]F=kx[/tex]

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]

where

m = 2.90 kg is the mass

[tex]v_{max}=1.05 m/s[/tex] is the maximum speed

Solving for E, we find

[tex]E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J[/tex]

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]

so the elastic potential energy is

[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J[/tex]

and since the total energy E = 1.60 J is conserved, the kinetic energy is

[tex]K=E-U=1.60 J-0.18 J=1.42 J[/tex]

And from the relationship between kinetic energy and speed, we can find the speed of the system:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s[/tex]

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]

So the restoring force exerted by the spring on the mass is

[tex]F=kx=(80 N/m)(0.0667 m)=5.34 N[/tex]

And so, we can calculate the acceleration by using Newton's second law:

[tex]a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2[/tex]

The spring constant of the dense object will be 80N/m.

The spring constant will be calculated thus:

k = f/m = 16/0.2 = 80N/m.

The frequency oscillation will be:

= 1/2π × ✓k/✓m

= 1/2π × ✓80/✓2.9

= 0.84 Hz

The maximum speed of the spring mass system will be:

v = [2π(0.84)] × 0.2

= 1.05m/s

The position on the x-axis where the maximum speed occur is at x = 0.

The maximum acceleration will be:

a = w²A

a = 5.25² × 0.2

= 5.51m/s²

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Answer:

P.E. = 6J

Explanation:

Hope this helps.  The answer above is incorrect.  It's 6.

To determine the potential energy at point C, we need to know the height difference between points A and C. The potential energy at point C can be calculated using the formula P.E. = mgh. Depending on the height difference given, the potential energy at point C would be either 16 joules or 18 joules.

To determine the potential energy at point C, we need to know the height difference between points A and C. The formula for potential energy is P.E. = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. If we assume the height difference is the same as in the given options, 4 or 6, we can calculate the potential energy at point C.

If the height difference is 4, then the potential energy at C would be 12 joules + 4 joules = 16 joules. If the height difference is 6, then the potential energy at C would be 12 joules + 6 joules = 18 joules.

Therefore, the potential energy at point C is either 16 joules or 18 joules, depending on the height difference between points A and C.

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(a) 25 A

The induced emf in the circuit is given by

[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]

where

[tex]\Delta Phi[/tex] is the variation of magnetic flux through the bracelet

[tex]\Delta t = 19.0 ms =0.019 s[/tex] is the time interval

The variation of magnetic flux is

[tex]\Delta \Phi = A \Delta B[/tex]

where

[tex]A=0.005 m^2[/tex] is the area enclosed by the bracelet

[tex]\Delta B=0.81 T-2.70 T=-1.89 T[/tex] is the change of magnetic field strength

So we find

[tex]\Delta \Phi = (0.005 m^2)(-1.89 T)=-9.45\cdot 10^{-3} Wb[/tex]

And so the induced emf is

[tex]\epsilon = -\frac{-9.45\cdot 10^{-3} Wb}{0.019 s}=0.50 V[/tex]

Since the resistance of the bracelet is

[tex]R=0.02\Omega[/tex]

the induced current is

[tex]I=\frac{V}{R}=\frac{0.50 V}{0.02 \Omega}=25 A[/tex]

(b) 12.5 W

The power delivered to the bracelet is given by

[tex]P=V I[/tex]

where we have

I = 25 A is the current

V = 0.50 V is the emf induced in the bracelet

Substituting numbers, we find

[tex]P=(0.50 V)(25 A)=12.5 W[/tex]

The problem involves using Faraday's Law to compute for the induced current and power in the bracelet due to a change in the magnetic field strength of the solenoid it is placed in.

This involves calculating the induced current (part a) and power (part b) in the bracelet due to a change in the magnetic field. This can be solved using Faraday's Law which states that the induced electromotive force (EMF) in a circuit is equal to the negative rate of change of magnetic flux. For part a, we compute for the EMF using the formula ΔΦ/Δt = (2.70 T - 0.81 T) * 0.00500 m²/0.019 s which can then be used to calculate the induced current by dividing the EMF by the resistance of the bracelet. Part b involves finding the delivered power using the formula P=I²R, where I is the induced current and R is the resistance.

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Answer:

[tex]4.8\cdot 10^{-15} N[/tex]

Explanation:

The force exerted on a charged particle by an electric field is:

[tex]F=qE[/tex]

where

q is the magnitude of the charge of the particle

E is the magnitude of the electric field

For an electron,

[tex]q=1.6\cdot 10^{-19} C[/tex]

while the magnitude of the electric field in the problem is

[tex]E=30000 V/m[/tex]

so, the electric force on the electron is

[tex]F=(1.6\cdot 10^{-19}C)(30000 V/m)=4.8\cdot 10^{-15} N[/tex]

(a) [tex]7.32\cdot 10^7 Hz[/tex]

The frequency of an electromagnetic waves is given by:

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

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

[tex]\lambda=4.1 m[/tex] is the wavelength of the wave in the problem

Substituting into the equation, we find

[tex]f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz[/tex]

(b) [tex]4.60\cdot 10^8 rad/s[/tex]

The angular frequency of a wave is given by

[tex]\omega = 2\pi f[/tex]

where

f is the frequency

For this wave,

[tex]f=7.32\cdot 10^7 Hz[/tex]

So the angular frequency is

[tex]\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s[/tex]

(c) [tex]1.53 m^{-1}[/tex]

The angular wave number of a wave is given by

[tex]k=\frac{2\pi}{\lambda}[/tex]

where

[tex]\lambda[/tex] is the wavelength of the wave

For this wave, we have

[tex]\lambda=4.1 m[/tex]

so the angular wave number is

[tex]k=\frac{2\pi}{4.1 m}=1.53 m^{-1}[/tex]

(d) [tex]1.03\cdot 10^{-6}T[/tex]

For an electromagnetic wave,

[tex]E=cB[/tex]

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

[tex]B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T[/tex]

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

[tex]I=\frac{E^2}{2\mu_0 c}[/tex]

where we have

E = 310 V/m is the amplitude of the electric field

[tex]\mu_0[/tex] is the vacuum permeability

c is the speed of light

Substituting into the formula,

[tex]I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2[/tex]

(g) [tex]1.53\cdot 10^{-8} kg m/s[/tex]

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

[tex]\frac{dp}{dt}=\frac{<S>A}{c}[/tex]

where the <S> is the magnitude of the Poynting vector, given by

[tex]<S>=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2[/tex]

and where the surface is

A = 1.8 m^2

Substituting, we find

[tex]\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s[/tex]

(h) [tex]8.47\cdot 10^{-7} N/m^2[/tex]

For a surface that totally absorbs the wave, the radiation pressure is given by

[tex]p=\frac{<S>}{c}[/tex]

where we have

[tex]<S>=254.2 W/m^2[/tex]

[tex]c=3\cdot 10^8 m/s[/tex]

Substituting, we find

[tex]p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2[/tex]

Answer:

1. From top to bottom and then back to top

2. From bottom to top and then back to bottom

Explanation:

As Time Period or periodic time period is time it takes to complete one complete cycle. So only these two options are correct. Yes ! If you assume a frictionless and isolated system then these two time intervals must be equal.

The motion of the suspended mass is simple harmonic motion; from the bottom position, then to equilibrium position, and then to the top position.

The period of a particle undergoing simple harmonic motion is defined as the time taken for the particle to complete one complete oscillation.

A mass suspended on a vertical spring and allowed to attain equilibrium. When, it is pulled downward and released, the mass begins to oscillate by moving up and down between the "top" and the "bottom" position.

The motion of the object is as follows:

The motion of the suspended mass is simple harmonic motion; from the bottom position, then to equilibrium position, and then to the top position.

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Answer:

speed =distance/time

Formula

s = d/t

s = speed

d = distance traveled

t = time elapsed

Thanks, Hope this helps.

Answer:

1440 W

Explanation:

First of all, we can find the total displacement of the sled, which is given by

[tex]d=\frac{1}{2}at^2[/tex]

where

a = 0.08 m/s^2 is the acceleration

t = 1 min = 60 s is the time

Substituting,

[tex]d=\frac{1}{2}(0.08 m/s^2)(60 s)^2=144 m[/tex]

Now we can find the wotk done on the sled, equal to the product between force and displacement:

[tex]W=Fd=(600 N)(144 m)=86,400 J[/tex]

And finally we can fidn the average power, which is the ratio between the work done and the time taken:

[tex]P=\frac{W}{t}=\frac{86,400 J}{60 s}=1440 W[/tex]

Answer:

b) Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

d) Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

f) Increasing the length of the wire will increase the resistance of the wire.

Answer:

option D

(2.6 , 3.1 )

Explanation:

First quadrant is where x and y values are positive.

Given in the question,

magnitude of vector = 4

angle at which vector is incline from x-axis = 50°

sin(50) = vertical / 4

vertical = sin(50)(4)

            = 3.06

            ≈ 3.1

  cosΔ = adj / hypo

cos(50) = horizontal / 4

horizontal = cos(50)(4)

                = 2.57

                ≈ 2.6

Answer:

A. A line can be drawn from the planet to the sun that sweeps out equal areas in equal times

Explanation:

This is exactly what Kepler's second law of planetary motion states:

"the segment joining the sun with the center of each planet sweeps out equal areas in equal time"

This law basically tells how the speed of a planet orbiting the sun changes during its revolution. In fact, we have that:

- when a planet is closer to the Sun, it will orbit faster

- when a planet is farther from the Sun, it will orbit slower

A) [tex]4.3\cdot 10^7 m/s[/tex]

For an electron moving in a region with both electric and magnetic field, the electron will move undeflected if the electric force on the electron is equal to the magnetic force:

[tex]qE= qvB[/tex]

which means that the speed of the electron will be

[tex]v=\frac{E}{B}[/tex]

where

E is the magnitude of the electric field

B is the magnitude of the magnetic field

In this problem,

[tex]B=2.5 mT=0.0025 T[/tex] is the intensity of the magnetic field

The electric field can be found as

[tex]E=\frac{V}{d}[/tex]

where

V = 600 V is the potential difference between the electrodes

[tex]d=5.6 mm=0.0056 m[/tex] is the distance between the electrodes

Substituting,

[tex]E=\frac{600 V}{0.0056 m}=1.07\cdot 10^5 V/m[/tex]

So the electron's speed is

[tex]v=\frac{1.07\cdot 10^5 V/m}{0.0025 T}=4.3\cdot 10^7 m/s[/tex]

B) [tex]9.8\cdot 10^{-2} m[/tex]

The radius of curvature of an electron in a magnetic field can be found by equalizing the centripetal force to the magnetic force:

[tex]m\frac{v^2}{r}=qvB[/tex]

where

m is the electron mass

v is the speed

r is the radius of curvature

q is the charge of the electron

Solving for r, we find

[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(4.3\cdot 10^7 m/s)}{(1.6\cdot 10^{-19} C)(0.0025 T)}=9.8\cdot 10^{-2} m[/tex]