The electric potential (voltage) at a specific location is equal to the potential energy per unit charge a charged object would have if it were at that location. If the zero point of the voltage is at infinity, the numerical value of the voltage is equal to the numerical value of work done to bring in a unit charge from infinity to that location. Select Values and Grid in the menu, and drag one positive charge to the middle of the screen, right on top of two intersecting bold grid lines. Using the voltage meter, you should find that 1 m away from the charge, the voltage is 9 V. What is the voltage 2 m away from the charge?

Answer:

The atomic number of a nucleus will go down by two in alpha decay

Answer:

The force is tripled too.

Explanation:

The magnitude of the electric force (repulsion or attraction) between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance that separates them (Coulomb's Law).

[tex]F=k_{e} \frac{[q_{1}][q_{2}]}{r^{2}}[/tex]

Where

[tex]k_{e}[/tex] is the Coulomb's constante.

[tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the signed magnitude of the charges.

[tex]r[/tex] the distance between charges.

So, if one of the charge is tripled, due the directly proportional product of the charges, the force will be tripled.

[tex]F=k_{e} \frac{[3q_{1}][q_{2}]}{r^{2}}[/tex]

or

[tex]F=k_{e} \frac{[q_{1}][3q_{2}]}{r^{2}}[/tex]

The result will be still the force tripled.

Answer:

-2 m/s

Explanation:

Assuming the collision is elastic, the total momentum must be conserved:

[tex]p_i = p_f[/tex]

[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]

where

m1 = 0.50 kg is the mass of the first ball

m2 = 1.00 kg is the mass of the second ball

u1 = +6.0 m/s is the initial velocity of the first ball

u2 = -12.0 m/s is the initial velocity of the second ball

v1 = -14 m/s is the final velocity of the first ball

v2 = ? is the final velocity of the second ball

By re-arranging the equation, we can find the final velocity of the 1.00 kg ball:

[tex]v_2 = \frac{m_1 u_1 + m_2 u_2 - m_1 v_1}{m_2}=\frac{(0.50 kg)(6.0 m/s)+(1.00 kg)(-12 m/s)-(0.50 kg)(-14 m/s)}{1.00 kg}=-2 m/s[/tex]

which means, 2 m/s in the same direction the second ball was travelling before the collision.

The speed of the second ball is 2 m/s

To calculate the speed of the second ball after the collision, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for v'

Note: The negative sign can be ignored in the final answer

Hence, the speed of the second ball is 2 m/s.

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Answer:

C

Explanation:

When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.

Now C has a charge of +12

When A and C come together they each have an equal share of that 12, so each of them has 6

So the answer is

A       B       C

6        0       6

which is C

In the equation for the magnetic field component of an electromagnetic wave, the quantities that appear include the electric field amplitude (E) and the speed of light (c), using the relationship B = E/c to relate the electric and magnetic field strengths.

The question pertains to the quantities that appear in the equation for the magnetic field component of an electromagnetic wave. In electrodynamics, the magnetic field component B of an electromagnetic wave can be calculated using the relationship “B = E/c”, where E is the electric field amplitude of the wave, and c is the speed of light in a vacuum, approximately 3 x 108 m/s. When a plane electromagnetic wave moves in a certain direction, its electric and magnetic fields oscillate perpendicular to the direction of wave propagation and to each other, a characteristic of a transverse wave. The strength of the magnetic field is directly proportional to the electric field and inversely proportional to the speed of light.

For example, if an electromagnetic wave has an electric field magnitude of 6.0 V/m, the magnitude of the magnetic field B can be found using the equation B = E/c. Similarly, for a more complex expression such as E = (6.0 × 10⁻³ V/m) sin(27 × 1¹/m - ωt), the associated magnetic field can be derived using the same principle, adjusted for the wave's specific amplitude, frequency, and phase.

Final answer:

The equation for the magnetic field component of an electromagnetic wave includes the electric field strength, the speed of light, and the permeability of free space.

Explanation:

The quantities that appear in the equation for the magnetic field component of an electromagnetic wave are the electric field strength (E), the speed of light (c), and the permeability of free space (μ₀). In the context of an electromagnetic wave traveling in a vacuum, the magnetic field component B can be expressed as B = E/c, assuming that both E and B are at right angles to the direction of propagation of the wave and to each other as per the transverse nature of electromagnetic waves. This relationship indicates that the magnetic field strength is directly proportional to the electric field strength and inversely proportional to the speed of light, highlighting the inherent interdependence between electric and magnetic fields in electromagnetic waves.

Thomson conducted a series of experiments on cathode ray tubes, which led him to the discovery of electrons.

This is how on April 30th, 1897 Thomson announced the discovery of the electron (although at that time he called it a corpuscle)

This discovery was considered one of the milestones in science of the late nineteenth century, since it allowed to have a new conception of the structure of matter and its interaction with energy.

It should be noted that from this discovery, Thomson proposed a new model of the atom that although erroneous, remained valid for years and served as the basis for new postulates and research on the subject.

Answer:

a

Explanation:

Answer:

0.122 m

Explanation:

We can imagine the hotspots in the microwave oven to correspond to the crests of the waves, so the distance between two hotspot is just the wavelength of the waves.

The wavelenght of an electromagnetic wave is given by

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

where

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

f is the frequency of the wave. In this case,

[tex]f=2.45 GHz = 2.45 \cdot 10^9 Hz[/tex]

so the wavelength is

[tex]\lambda=\frac{3\cdot 10^8 m/s}{2.45\cdot 10^9 Hz}=0.122 m[/tex]

The separation between hotspots in a 2.45 GHz microwave oven is approximately half the wavelength of the microwaves, which is around 6.1 cm. This distance is derived from the speed of light and the frequency of the microwaves used.

The separation between hotspots in a microwave oven can be determined by understanding the wavelength of the microwaves used, which for most household microwave ovens is set at a frequency of 2.45 GHz. When these microwaves are generated within the oven, they reflect off the interior walls, creating standing waves. At points where the crests or troughs of these microwaves align, hotspots are formed due to the concentration of energy.

This heating effect occurs because microwaves at a frequency of 2.45 GHz have the right amount of energy to cause polar molecules, like water, to rotate, thereby transferring kinetic energy into thermal energy. The dipole nature of water molecules allows for efficient energy transfer, specifically heating the areas where these molecules are present, such as in food.

To find the separation between the hotspots, we use the relationship between the speed of light (c), frequency (f), and wavelength (λ) given by c = λ * f. With c being approximately 3 x 108 meters per second and f being 2.45 GHz, we can calculate the wavelength and thus the distance between hotspots. The speed of the microwaves is the speed of light, and so the wavelength (λ) for 2.45 GHz radiation would be about 0.122 meters (12.2 cm), hence the hotspots would be separated by approximately half a wavelength, around 6.1 cm.

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a) 8.58 m

For an open-open tube, the fundamental frequency is given by

[tex]f=\frac{v}{2L}[/tex]

where

f is the fundamental frequency (the lowest frequency)

v is the speed of sound

L is the length of the tube

In this problem, we have

f = 20 Hz

v = 343 m/s (speed of sound in air)

Solving the equation for L, we find the shortest length of the tube:

[tex]L=\frac{v}{2f}=\frac{343 m/s}{2(20 Hz)}=8.58 m[/tex]

(b) 4.29 m

For an open-closed tube, the fundamental frequency is instead given by

[tex]f=\frac{v}{4L}[/tex]

Where in this problem, we have

f = 20 Hz

v = 343 m/s (speed of sound in air)

Solving the equation for L, we find the shortest length of the tube:

[tex]L=\frac{v}{4f}=\frac{343 m/s}{4(20 Hz)}=4.29 m[/tex]

Answer:

2.5s

Explanation:

Given parameters:

Decay constant of the nuclide = 4.6 x 10⁻³s⁻¹

Unkown:

Half life =?

The half life is the time take for half of the radioactive nuclei to disintegrate. It is mathematically expressed as:

Half life = 0.693/λ

Where λ is the decay constant

Since we have been given λ we plug it into the equation:

Half life = 0.693/4.6 x 10⁻³

Half life = 0.693/0.0046

Half life =150. 65s

Now we convert to minutes because the options are in minutes:

60s = 1 minute

150.65s = 150.65/60 = 2.5s

The half life of the nuclide is 2.5s

Answer:

20.The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is

Explanation:

1) [tex]2.43\cdot 10^{-9} C[/tex]

The electric field produced by a small charged sphere outside sphere is identical to that of a single point charge:

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

where

k is the Coulomb constant

Q is the charge on the sphere

r is the distance from the sphere

Here we have

E = 350 N/C is the electric field

r = 0.25 m is the distance from the sphere

Solving for Q, we find the charge:

[tex]Q=\frac{Er^2}{k}=\frac{(350 N/C)(0.25 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=2.43\cdot 10^{-9} C[/tex]

2) 1.29 N

First of all, we need to find the magnitude of the electric field produced by the charge:

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

where here we have

[tex]Q=3.6\cdot 10^{-6}C[/tex] is the charge source of the field

r = 0.23 m is the distance at which we have to calculate the field

Substituting,

[tex]E=(9\cdot 10^9 Nm^2 C^{-2}) \frac{3.6\cdot 10^{-6}C}{(0.23 m)^2}=6.12\cdot 10^5 N/C[/tex]

Now we can calculate the electric force exerted on the second charge, which is given by

[tex]F=qE[/tex]

where

[tex]q=2.1\cdot 10^{-6} C[/tex] is the magnitude of the second charge

Substituting,

[tex]F=(2.1\cdot 10^{-6} C)(6.12\cdot 10^5 N/C)=1.29 N[/tex]

3) 3.33 W

The power generated in a circuit is given by:

P = V I

where

V is the voltage

I is the current

In this circuit, we have

V = 9 V

I = 0.37 A

So the power generated is

[tex]P=(9 V)(0.37 A)=3.33 W[/tex]

4) [tex]+1.1\cdot 10^{-7}C[/tex]

The magnitude of the force between two electric charge is given by Coulomb's law:

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

where q1 and q2 are the two charges, and r is their separation.

In this problem we have

[tex]q_1 = 6.0\cdot 10^{-6}C[/tex]

[tex]F=15 N[/tex]

[tex]r=0.02 m[/tex]

Solving for q2, we find the magnitude of the second charge

[tex]q_2 = \frac{Fr^2}{k q_1}=\frac{(15 N)(0.02 m)^2}{(9\cdot 10^9 Nm^2 C^{-2})(6.0\cdot 10^{-6}C)}=1.1\cdot 10^{-7}C[/tex]

Moreover, the force is attractive: this means that the two charges must have opposite signs (in fact, like charges repel each other, while unline charges attract each other). In this problem, q1 is negative, so q2 must be positive.

5a) -0.07 N (towards negative x-direction)

First of all, we need to calculate the force exerted on sphere A by sphere B; this is given by

[tex]F_{AB} = k \frac{q_A q_B}{r_{AB}^2}[/tex]

where

[tex]q_A = 2.0\cdot 10^{-6}C\\q_B = -3.6\cdot 10^{-6}C\\r_{AB}=0.6 m[/tex]

Substituting,

[tex]F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N[/tex]

(the negative sign means the force is towards the left)

The force exerted on sphere A by sphere C is

[tex]F_{AC} = k \frac{q_A q_C}{r_{AB}^2}[/tex]

where

[tex]q_A = 2.0\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{AC}=0.8 m[/tex]

Substituting,

[tex]F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N[/tex]

So the net force on sphere A is

[tex]F=F_{AB}+F_{AC}=-0.18 N+0.11 N=-0.07 N[/tex]

5b) +3.06 N (towards positive x-direction)

The force exerted on sphere B by sphere A has been calculate at step 5a), and it is

[tex]F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N[/tex]

and the force is attractive, so towards the left

The force exerted on sphere B by sphere C is

[tex]F_{BC} = k \frac{q_B q_C}{r_{BC}^2}[/tex]

where

[tex]q_B = -3.6\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{BC}=0.8 m-0.6 m=0.2 m[/tex]

Substituting,

[tex]F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N[/tex]

and the force is still attractive, so towards the right

So the net force on sphere B is

[tex]F=F_{AB}+F_{BC}=-0.18 N+3.24 N=3.06 N[/tex]

5c) -3.13 N (towards negative x-direction)

The force exerted on sphere C by sphere A has been calculate at step 5a), and it is

[tex]F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N[/tex]

and the force is repulsive, so towards the right

Similarly, we found the force exerted on sphere C by sphere B at step 5b)

[tex]F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N[/tex]

and the force is attractive, so towards the left

So the net force on sphere C is

[tex]F=F_{AC}+F_{BC}=+0.11+(-3.24 N)=-3.13 N[/tex]

To calculate the number of photons emitted in one flash, use the formula: Number of photons = Energy of the flash / Energy of one photon. By using the given power and duration of the flash, you can calculate the energy of the flash and then calculate the energy of one photon using the wavelength. Finally, divide the energy of the flash by the energy of one photon to find the number of photons emitted.

To calculate the number of photons emitted in one flash, we need to use the formula:

Number of photons = Energy of the flash / Energy of one photon

First, let's calculate the energy of the flash. The power of the flash is given as 1.2 mW, and it lasts for 100 ms. Using the formula P = E/t, we can find the energy of the flash as 1.2 mJ.

Next, we need to calculate the energy of one photon. The energy of a photon can be calculated using the formula E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.

The wavelength is given as 550 nm, which is equivalent to 550 x 10^-9 m. Plugging in the values, we can calculate the energy of one photon as 3.606 x 10^-19 J.

Now, we can use the formula to calculate the number of photons:

Number of photons = 1.2 x 10^-3 J / 3.606 x 10^-19 J = 3.328 x 10^15 photons

Therefore, approximately 3.328 x 10^15 photons are emitted in one flash.

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To calculate the number of photons emitted in one flash, we can use the relationship between power and energy. The energy of a single photon can be determined using the wavelength. The number of photons emitted in one flash is approximately 3.33 x 10^14 photons.

To calculate the number of photons emitted in one flash, we need to use the relationship between power and energy of a photon. Power is given as 1.2 mW and since the flash lasts for 100 ms, we can calculate the energy of the flash as:
E = P x t = (1.2 mW) x (100 ms) = 0.12 mJ

To find the number of photons, we can use the formula:
Number of photons = Energy of flash / Energy of a single photon

The energy of a single photon can be determined using the relationship between energy and wavelength:
Energy = hc / λ

Where h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the values:
Energy = (6.63 x 10^-34 J.s) x (3 x 10^8 m/s) / (550 x 10^-9 m) ≈ 3.6 x 10^-19 J

Finally, plugging the values into the formula:
Number of photons = (0.12 x 10^-3 J) / (3.6 x 10^-19 J) ≈ 3.33 x 10^14 photons

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Given:

u = 8.5 cm

f = -14.0 cm

v = image distance

Using the mirror formula 1/u + 1/v = 1/f

1/8.5 + 1/v = 1/-14.0

Rewrite to solve for v:

v = (8.5 * -14.0) / (8.5 - (-14.0))

v = -119 / 22.5

v = -5.29 cm  ( round answer as needed.)

Speed because all waves in the electromagnetic spectrum travel at the speed of light

Answer:

b) It is impossible to tell without knowing the masses.

Explanation:

The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by

[tex]\Delta T= \frac{Q}{m C_s}[/tex]

where

Q is the amount of heat

m is the mass of the substance

Cs is the specific heat capacity of the substance

In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.

Answer:

116

Explanation:

The transformer equation states that:

[tex]\frac{V_p}{V_s}=\frac{N_p}{N_s}[/tex]

where

Vp is the voltage in the primary coil

Vs is the voltage in the secondary coil

Np is the number of turns in the primary coil

Ns is the number of turns in the secondary coil

Here we have:

Vp = 5670 V

Np = 2750

Vs = 240 V

So we can solve the formula to find Ns:

[tex]N_s = N_p \frac{V_s}{V_p}=(2750)\frac{240 V}{5670 V}\sim 116[/tex]

So, the secondary coil has 116 turns.

"ps" stands for "pico-second".  It means 10⁻¹² second.

The period of this wave is 8.43 x 10⁻¹² second.

In a vacuum, the speed of any electromagnetic wave is 3 x 10⁸ m/s.

Wavelength = (speed) / (frequency)

But Frequency = 1 / period

So Wavelength = (speed) x (period)

Wavelength = (3 x 10⁸ m/s) x (8.43 x 10⁻¹² sec)

Wavelength = 2.5 x 10⁻³ meter

Wavelength = 2.5 millimeters

You can find the wavelength of an electromagnetic wave with a given period by first calculating frequency using the formula f = 1/T. Then, use the universal wave equation c = λv, to find the wavelength, λ = c/f. Plugging all the values, we find the wavelength is about 2.53 meters.

The subject of this question revolves around an

electromagnetic wave

with a given oscillation period and we need to find its wavelength. To solve this, we first have to convert the period into frequency, as the two are inverse to each other.

The frequency (f) can be calculated by taking the reciprocal of the period (T), so f = 1/T. Given that T = 8.43 ps (ps is picoseconds and 1 ps equals 10^-12 seconds), the frequency would therefore be f = 1/(8.43x10^-12 s) which equals approximately 1.19x10^11 Hz (Hz is the unit of frequency).

Now we use the universal wave equation, c = λv, where c is the speed of light (approximately 3 x 10^8 ms^-1 in a vacuum), v is the frequency, and λ is the wavelength we're trying to find. Solving for λ, we get λ = c/f. Substituting the values, we get λ = (3x10^8 m/s) / (1.19x10^11 Hz), which computes to approximately λ = 2.53m. Therefore, the wavelength of this electromagnetic wave is approximately 2.53 meters.

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Answer:

radar waves

Explanation:

Wavelength chart.

Infrared is barely outside of the visible spectrum, but radar is (comparatively) way outside the visible spectrum on the side of longer wavelength.

Longest . . . radar waves

then . . . infrared

then . . . ultraviolet

then . . . X-rays

Shortest . . . gamma rays

Conservation of energy means that initial kinetic energy =maximum gravitational potential energy. 1/2 * 3 * 10^2 = 150J. The gravitational potential energy at the top is 150 joules. The gravitational potential energy is calculated through mgh, so 150 = 3 * 10 * h. Height is 5 meters

Answer: isotopes

An isotope is an atom of the same chemical element that has the same number of protons in its nucleus, but not the same number of neutrons.

Therefore, this isotope atom will have the same atomic number and will be located in the same place of the periodic table of the main chemical element, but different mass number.

In other words, the isotopes of a chemical element have a different mass although they have the same atomic number, which is directly related to the number of protons in the nucleus of the atom.

Final answer:

Atoms of the same element that possess different numbers of neutrons are known as isotopes, and these isotopes have the same atomic number but a different mass number.

Explanation:

Atoms of the same element with different numbers of neutrons are called isotopes. For instance, within hydrogen, there are notable examples like deuterium and tritium, which respectively have one and two neutrons, while the most common hydrogen isotope has none. Isotopes of an element share the same atomic number due to having an identical number of protons but possess a different mass number, which is the sum of protons and neutrons in the atom's nucleus.

Answer:

Increase, increase

Explanation:

For a light bulb, voltage, resistance and current follows the Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage

R is the resistance

I is the current

In this problem, the voltage is fixed: V=120 V. We can rewrite the equation as

[tex]I=\frac{V}{R}[/tex] (1)

we see that the current is inversely proportional to the resistance: so, if we decrease the resistance of the filament, the current will increase.

Moreover, the power dissipated by the bulb is

[tex]P=I^2 R[/tex]

which can be rewritten using (1) as

[tex]P=\frac{V^2}{R}[/tex]

We see that the power dissipated is inversely proportional to the resistance: therefore, if we decrease the resistance of the filament, the power dissipated will increase.

Answer:

Explanation:

Good thing it says approximately. Your best guess is just a guess. It takes 1 hz (or 1 sine wave) about 0.62 seconds before it begins to repeat itself.

The frequency is  # of cycles / time = 1 hz / 0.62 = 1.61 hz/second.

Answer:

Let's start by explaining that each atom in its natural state has a specific structure of its energy levels. Where the lowest energy level is called the ground state.

At this level, as long as no energy is communicated to the atom, the electron will remain in the ground state.

What does this mean?

When an atom is in its ground state, its electrons fill the lower energy orbitals completely before they begin to occupy higher energy orbitals.

If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, the far of b if a is 200 mpc away is   100 Mpc

The parsec ([tex]pc[/tex]) is a unit of length used to measure the large distances to astronomical objects outside the Solar System.

Although parsecs are used for the shorter distances within the Milky Way, multiples of parsecs are also required for the larger scales in the universe such as Kiloparsecs (kpc) is for the more distant objects within and around the Milky Way, megaparsecs (Mpc) for mid-distance galaxies, and gigaparsecs (Gpc) for many quasars and the most distant galaxies.

If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, how far away is b if a is 200 mpc away? Remember that [tex]v = h_0 * d[/tex]

galaxy A:  [tex]2(galaxy B's V) = H_0 * 200[/tex]

[tex]galaxy B's V = H_0 * 100[/tex]

galaxy B:  100 Mpc

Grade:  9

Subject:  physics

Chapter:  the recession velocities

Keywords:  galaxy,  the recession velocities, velocity, mpc, measure, far away

static electricity and friction

Answer:

1. line of best fit---(a line that best represents the trend of the data)

2. slope---(the change in the y variable compared to the change in the x variable between two points on a graph)

3. variable---(a quantity that changes during an experiment)

Explanation:

Technically the variable one could be considered wrong, as some variables stay the same, but otherwise these are the answers.

Answer:

0.0129 V

Explanation:

The average emf induced in the coil is given by Faraday-Newmann-Lenz:

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

where

[tex]\Delta \Phi_B[/tex] is the variation of magnetic flux through the coil

[tex]\Delta t = 15 s[/tex] is the time interval

Here we have

B = 0.50 T is the strength of the magnetic field

The radius of the coil is

r = 35 cm = 0.35 m

So the area is

[tex]A=\pi r^2 = \pi (0.35 m)^2=0.385 m^2[/tex]

The initial flux through the coil is

[tex]\Phi_i = BA = (0.50 T)(0.385 m^2)=0.193 Wb[/tex]

while the final flux is zero, since the coil has been completely removed from the magnetic field region; so, the variation of magnetic flux is

[tex]\Delta \Phi = \Phi_f - \Phi_i = -0.193 Wb[/tex]

and so, the average emf induced is

[tex]\epsilon=-\frac{-0.193 Wb}{15 s}=0.0129 V[/tex]

D) It is a good conductor of electricity

The correct answer is B) the combined mass of two particles is completely transformed into energy (photons).