Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00μm. The electrons then head toward an array of detectors a distance 1.068 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.527 cm from the center of the pattern. What is the wavelength λ of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=Lλ/a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves.

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:

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: 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.

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:

A. The direction of the force and the direction of acceleration must be the same as each other.

Explanation:

Force can be defined as push or pull. An unbalanced force that is non-zero net force causes a body to accelerate. Newton's second law states that acceleration depends on the force.

F = m a

where m is the mass of the body and a is the acceleration.

Increase in force causes increase in acceleration. The direction of acceleration and direction of force are same.

Considering acceleration due to gravity and force of acceleration - gravitational force always acts along the line joining the centers of two bodies and so, the direction of the acceleration due to gravity also is in the same direction.

Answer:

The direction of the force and the direction of acceleration must be the same as each other.

Explanation:

According to Newton's second law of motion, the force acting on an object  depends on its acceleration. Its formula is given by :

F = m a

Where,

m is the mas of an object

a is its acceleration.

i.e. force is directly proportional to the acceleration of an object. The direction of force and acceleration must be same.  

Hence, the correct option is (A) "The direction of the force and the direction of acceleration must be the same as each other".

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.

Answer:

speed =distance/time

Formula

s = d/t

s = speed

d = distance traveled

t = time elapsed

Thanks, Hope this helps.

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.

What area ?

What column of water ?

The force of gravity acting on a column of water with an area of 2 square meters and height of 1 meter is 19620 Newtons. This calculation is based on the assumption of the density of water being 1000 kg/m³ and the acceleration due to gravity being 9.81 m/s².

The force of gravity acting on a column of water with an area of 2 square meters can be calculated using the concepts of fluid pressure and weight. First, you need to understand that the force of gravity on the water can be expressed by the formula Weight = Mass x Gravity. The Mass can be derived from the Volume (Area x Height) and the density of the water (approximately 1000 kg/m³).

Thus, assuming the column of water is 1m high, the volume of water is 2 m² x 1 m = 2 m³. The mass is then Volume x Density = 2 m³ x 1000 kg/m³ = 2000 kg. Substituting these values into the weight formula gives us Weight = 2000 kg x 9.81 m/s² (acceleration due to gravity), which equals 19,620 Newtons.

Therefore, the force of gravity acting on the column of water is 19620 N.

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(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

[tex]V = V_R + V_L[/tex]

The potential difference across the inductance is given by

[tex]V_L(t) = V e^{-\frac{t}{\tau}}[/tex] (1)

where

[tex]\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s[/tex] is the time constant of the circuit

At time t=0,

[tex]V_L(0) = V e^0 = V = 20 V[/tex]

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

[tex]V_R = V-V_L = 20 V-20 V=0[/tex]

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

[tex]t >> \tau[/tex]

Since [tex]\tau[/tex] is at the order of less than milliseconds.

Using eq.(1), we see that for [tex]t >> \tau[/tex], the exponential becomes zero, and therefore the potential difference across the coil is zero:

[tex]V_L = 0[/tex]

Therefore, the potential difference across the resistor will be

[tex]V_R = V-V_L = 20 V- 0 = 20 V[/tex]

(c) Yes

The two voltages will be equal when:

[tex]V_L = V_R [/tex] (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

[tex]V=V_L +V_R[/tex]

And rewriting this equation,

[tex]V_R = V-V_L[/tex]

Substituting into (2) we find

[tex]V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V[/tex]

So, the two voltages will be equal when they are both equal to 10 V.

(d) at [tex]t=5.77\cdot 10^{-4}s[/tex]

We said that the two voltages will be equal when

[tex]V_L=\frac{V}{2}[/tex]

Using eq.(1), and this last equation, this means

[tex]V e^{-\frac{t}{\tau}} = \frac{V}{2}[/tex]

And solving the equation for t, we find the time t at which the two voltages are equal:

[tex]e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s[/tex]

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

[tex]I_0 = 3.20 A[/tex]

The problem says this current is stable: this means that we are in a situation in which [tex]t>>\tau[/tex], so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

[tex]V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V[/tex]

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

[tex]V_L(0)=.V[/tex]

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

[tex]t>>\tau[/tex]

If we use this approximation into the formula

[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)

We find that

[tex]V_L = 0[/tex]

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

When a battery is connected to a coil and a resistor, initially all the battery voltage is across the coil. After some time, the voltage across the coil reduces to zero and that across the resistor becomes equal to the battery voltage. The voltages are equal during the magnetization process of the coil. When the battery is replaced with a short circuit, the entire voltage falls across the coil initially and then reduces to zero.

At the instant the battery is connected to the 5.00mH coil and 6.00Ω resistor (t=0), the potential difference across the resistor will be zero and the entire battery voltage will be across the coil as it opposes the sudden change in current. Therefore, at t=0, the voltage across the resistor is 0V and the voltage across the coil is 20.0V.

After several seconds, the current in the circuit will have achieved a steady state as the coil has become fully magnetized and now behaves as a wire. Therefore, the voltage drop across the resistor will become 20.0V (due to Ohm's law I=V/R, V=IR), and that across the coil will be 0V.

The two voltages are equal when the coil is half-way through its magnetization process. This is when the current in the circuit is building up.

When the circuit is modified with a short circuit after a current of 3.20A has been established, the entire potential difference falls across the coil immediately afterwards. Several seconds later, the coil loses its magnetic field since no more current flows through the circuit, thus no voltage exists either across the coil or the resistor.

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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. [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:

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:

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:

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.

A) [tex]7.8\cdot 10^{-10} N[/tex]

First of all, we need to find the velocity of the ball as it enters the magnetic field region.

Since the ball starts from a height of h=110 m and has a vertical acceleration of g=9.8 m/s^2 (acceleration due to gravity), the final velocity can be found using the equation

[tex]v^2 = u^2 +2gh[/tex]

where u=0 is the initial velocity. Solving for v,

[tex]v=\sqrt{2(9.8 m/s^2)(110 m)}=46.4 m/s[/tex]

The ball contains

[tex]N=4.20\cdot 10^8[/tex] excess electrons, each of them carrying a charge of magnitude [tex]e=1.602\cdot 10^{-19}C[/tex], so the magnitude of the net charge of the ball is

[tex]Q=Ne=(4.20\cdot 10^8)(1.602\cdot 10^{-19}C)=6.72\cdot 10^{-11}C[/tex]

And given the magnetic field of strength B=0.250 T, we can now find the magnitude of the magnetic force acting on the ball:

[tex]F=qvB=(6.72\cdot 10^{-11}C)(46.4 m/s)(0.250 T)=7.8\cdot 10^{-10} N[/tex]

B) From north to south

The direction of the magnetic force can be found by using the right-hand rule:

- index finger: direction of motion of the ball -> downward

- middle finger: direction of magnetic field --> from east to west

- thumb: direction of the force --> from south to north

However, this is valid for a positive charge: in this problem, the ball is negatively charged (it is made of an excess of electrons), so the direction of the force is reversed: from north to south.

The magnitude of the force that the magnetic field exerts on the ball is 7.812 x 10⁻¹⁰ N.

The magnetic force will directed north to south (reverse direction due to charge of ball).

The given parameters;

The speed of the charge is calculated by applying the principle of conservation of mechanical energy;

Potential energy at top = kinetic energy at bottom

mgh = ¹/₂mv²

2gh = v²

[tex]v = \sqrt{2gh} \\\\v= \sqrt{2\times 9.8 \times 110} \\\\v = 46.43 \ m/s[/tex]

The charge of the excess electron is calculated as follows;

Q = Ne

[tex]Q = (4.2 \times 10^8)\times (1.602 \times 10^{-19})\\\\Q = 6.73 \times 10^{-11} \ C[/tex]

The magnitude of the force that the magnetic field exerts on the ball is calculated as follows;

[tex]F = qvB\\\\F = 6.73 \times 10^{-11} \times 46.43 \times 0.25\\\\F = 7.812 \times 10^{-10} \ N[/tex]

The direction of the magnetic force will be perpendicular to speed of the charge and magnetic field.

The magnetic field is directed to east, the magnetic force will directed south to north. Since the ball is negatively charged (excess electron), the direction will be reversed. Thus, the force will be directed from north to south.

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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.

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]

(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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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 :)

(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.

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.

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:

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]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]

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:

(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.

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