The speed of the electrons in the wire is 2.44 × 106 m/s.2. The time interval over which a 0.133-mm layer of silver builds up on the teapot is 7.52 hours.3a.
The drift speed of the electrons in the copper wire is 2.29 × 10-5 m/s.3b. The drift speed of electrons increases as the number of conduction electrons per atom increases. 4a. The current in the lightbulb is 0.744 A.4b. Short Answer: The current in the lightbulb would be 0.930 A if it were used in one, where the potential difference across it would be 220 V.5. Short Answer: The current in the copper wire is 2.73 A.
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What is the total energy of a 0.90 g particle with a speed of 0.800? Express your answer in joules to two significant figures.
The total energy of a 0.90 g particle with a speed of 0.800 m/s is 0.036 J.
The total energy of a particle can be calculated using the formula: Total energy = Kinetic energy
The kinetic energy of a particle is given by the formula: Kinetic energy = (1/2) * mass * speed²
First, we need to convert the mass of the particle from grams to kilograms: Mass = 0.90 g = 0.90 * 10⁻³ kg = 9.0 * 10⁻⁴ kg
Next, we can substitute the values into the formula for kinetic energy: Kinetic energy = (1/2) * (9.0 * 10⁻⁴ kg) * (0.800 m/s)²
Simplifying the expression: Kinetic energy = (1/2) * (9.0 * 10⁻⁴) * (0.800)²
Kinetic energy = 3.60 * 10⁻⁴ J
Rounding the answer to two significant figures: Kinetic energy = 0.036 J
Therefore, the total energy of the particle is 0.036 J to two significant figures.
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A rock is dropped at time t=0 from a tower 50−m high. 1 second later a second rock is thrown downward from the same height. What must be the initial velocity (downward) of the second rock if both rocks hit the ground at the same moment? 15.4 m/s 9.8 m/s 12 m/s 16 m/s
The initial velocity (downward) of the second rock must be approximately 101 m/s if both rocks hit the ground at the same moment.
We are given that a rock is dropped at time t = 0 from a tower 50 m high. One second later, a second rock is thrown downward from the same height. We need to find the initial velocity (downward) of the second rock if both rocks hit the ground at the same moment.
Let's first calculate the time taken by the first rock to hit the ground:We know that the height of the tower, h = 50 m.Let g = 9.8 m/s² be the acceleration due to gravity.
As the rock is being dropped, its initial velocity u is zero.Let the time taken by the first rock to hit the ground be t₁.
Using the formula: h = ut + (1/2)gt² ,
50 = 0 + (1/2) * 9.8 * t₁²,
0 + (1/2) * 9.8 * t₁² ⇒ t₁ = √(50 / 4.9) ,
t₁ = 3.19 s.
Now let's consider the second rock. Let its initial velocity be u₂.The time taken by the second rock to hit the ground is
t₁ = t₁ - 1 ,
t₁ - 1 = 2.19 s.
We know that the acceleration due to gravity is g = 9.8 m/s².Using the formula: h = ut + (1/2)gt²
50 = u₂(2.19) + (1/2) * 9.8 * (2.19)².
u₂(2.19) + (1/2) * 9.8 * (2.19)²⇒ 245 ,
245 = 2.19u₂ + 22.9,
2.19u₂ + 22.9⇒ 2.19u₂,
2.19u₂= 222.1,
u₂ = 222.1 / 2.19,
u₂ ≈ 101.37,
u₂ ≈ 101 m/s.
Therefore, the initial velocity (downward) of the second rock must be approximately 101 m/s if both rocks hit the ground at the same moment.
Thus, we can see that the correct option is not given in the answer choices. The correct answer is 101 m/s.
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15) During a 4.50 s time period the magnetic field through a 0.350 m² wire loop changes from 2.30 T to 5.50T (directed straight through the loop), what is the average induced emf in the wire? 4.sos & ang NAER • 6.350m2
Given a change in magnetic field from 2.30 T to 5.50 T over a time period of 4.50 s, and a wire loop with an area of 0.350 m²,The average induced emf in the wire loop is 5.33 V.
According to Faraday's law, the induced emf in a wire loop is equal to the rate of change of magnetic flux through the loop. The magnetic flux (Φ) is given by the product of the magnetic field (B) and the area of the loop (A). In this case, the magnetic field changes from 2.30 T to 5.50 T, so the change in magnetic field (ΔB) is 5.50 T - 2.30 T = 3.20 T.
The average induced emf (ε) can be calculated using the formula:
ε = ΔΦ / Δt
where ΔΦ is the change in magnetic flux and Δt is the change in time. The change in time is given as 4.50 s.
To find the change in magnetic flux, we multiply the change in magnetic field (ΔB) by the area of the loop (A):
ΔΦ = ΔB * A
Plugging in the values, we have:
ΔΦ = 3.20 T * 0.350 m² = 1.12 Wb (weber)
Finally, substituting the values into the formula for average induced emf, we get:
ε = 1.12 Wb / 4.50 s = 5.33 V
Therefore, the average induced emf in the wire loop is 5.33 V.
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The potential at the surface of a sphere (radius R) is given by Vo = k cos (30), where k is a constant. a) Find the potential inside the sphere. (5 points) b) Find the potential outside the sphere. (5 points) c) Calculate the surface charge density o(0). (5 points)
Surface charge density σ0 on the surface of the sphere is given by σ0 = ε0(k√3/2 - k/2R).
Given that the potential at the surface of a sphere (radius R) is given by Vo=k cos(30), where k is a constant. Our task is to find the potential inside the sphere, and the potential outside the sphere, and calculate the surface charge density σ0(a).
a) Find the potential inside the sphere
The potential inside the sphere is given by;
V(r) = kcos(30)×(R/r)
On substituting the given value of k and simplifying, we get:
V(r) = (k√3/2)×(R/r)
Potential inside the sphere is given by V(r) = (k√3/2)×(R/r).
b) Find the potential outside the sphere
The potential outside the sphere is given by;
V(r) = kcos(30)×(R/r²)
On substituting the given value of k and simplifying, we get;
V(r) = (k/2)×(R/r²)
Potential outside the sphere is given by V(r) = (k/2)×(R/r²).
c) Calculate the surface charge density o(0)
Surface charge density on the surface of the sphere is given by;
σ0 = ε0(E1 - E2)
On calculating the electric field inside and outside the sphere, we get;
E1 = (k√3/2)×(1/R) and
E2 = (k/2)×(1/R²)σ0
= ε0[(k√3/2)×(1/R) - (k/2)×(1/R²)]
On substituting the given value of k and simplifying, we get;
σ0 = ε0(k√3/2 - k/2R)
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A series RLC circuit has a resistor and an inductor of known values (862 Ω and 11.8mH, respectively) but the capacitance C of the capacitor is unknown. To find its value, an ac voltage that peaks at 50.0 V is applied to the circuit. Using an oscilloscope, you find that resonance occurs at a frequency of 441 Hz. In μF, what must be the capacitance of the capacitor?
The capacitance of the capacitor in the RLC circuit must be approximately 1.51 μF.
To find the capacitance of the capacitor in the RLC circuit, we can use the resonance condition. At resonance, the inductive reactance and capacitive reactance cancel each other out, resulting in a purely resistive impedance.The resonance frequency (fr) of the circuit is given as 441 Hz. At resonance, the inductive reactance (XL) and capacitive reactance (XC) can be calculated using the following formulas: XL = 2πfL
XC = 1 / (2πfC)Since XL = XC at resonance, we can equate these two equations:
2πfL = 1 / (2πfC)
Simplifying the equation:
2πfL = 1 / (2πfC)
2πfC = 1 / (2πfL)
C = 1 / (4π²f²L)
Substituting the given values:
C = 1 / (4π² * (441 Hz)² * (11.8 mH))
Converting 11.8 mH to farads:
C = 1 / (4π² * (441 Hz)² * (11.8 × 10⁻³ H))
C ≈ 1.51 μF
Therefore, the capacitance of the capacitor in the RLC circuit must be approximately 1.51 μF.
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The idea that force causes acceleration doesn’t seem strange. This and other ideas of Newtonian mechanics are consistent with our everyday experience. Why do the ideas of relativity seem strange? 1. The effects of relativity become apparent only at very high speeds very uncommon to everyday experience. 2. Earth’s rotation doesn’t let us observe relativity that applies to systems moving in straight trajectories. 3. The principles of relativity apply outside Earth. 4. For the effects of relativity to become apparent large masses are needed.
The ideas of relativity seem strange compared to Newtonian mechanics because their effects are only apparent at very high speeds, which are uncommon in everyday experience. Earth's rotation also limits our ability to observe relativity, as it applies to systems moving in straight trajectories. Additionally, the principles of relativity extend beyond Earth and apply in various scenarios. Lastly, the effects of relativity become more pronounced with large masses. These factors contribute to the perception that the ideas of relativity are unfamiliar and counterintuitive.
The principles of relativity, as formulated by Albert Einstein, can appear strange because their effects are most noticeable at speeds that are far beyond what we encounter in our daily lives. Relativity introduces concepts like time dilation and length contraction, which become significant at velocities approaching the speed of light. These speeds are not typically encountered by humans, making the effects of relativity seem abstract and distant from our everyday experiences.
Earth's rotation further complicates our ability to observe relativity's effects. Relativity primarily applies to systems moving in straight trajectories, while Earth's rotation introduces additional complexities due to its curved path. As a result, the apparent effects of relativity are not easily observable in our day-to-day lives.
Moreover, the principles of relativity extend beyond Earth and apply in various scenarios throughout the universe. The behavior of objects, the passage of time, and the properties of light are all influenced by relativity in a wide range of cosmic settings. This universality of relativity contributes to its seemingly strange nature, as it challenges our intuitive understanding based on Earth-bound experiences.
Lastly, the effects of relativity become more pronounced with large masses. Gravitational fields, which are described by general relativity, become significant around massive objects like stars and black holes. Consequently, the predictions of relativity become more evident in these extreme environments, where the warping of spacetime and the bending of light can be observed.
In summary, the ideas of relativity appear strange compared to Newtonian mechanics due to the combination of their effects being noticeable only at high speeds, limited observations caused by Earth's rotation, the universal application of relativity, and the requirement of large masses for the effects to become apparent. These factors contribute to the perception that relativity is unfamiliar and counterintuitive in our everyday experiences.
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Suppose that a parallel-plate capacitor has circular plates with radius R = 39 mm and a plate separation of 3.9 mm. Suppose also that a sinusoidal potential difference with a maximum value of 180 V and a frequency of 75 Hz is applied across the plates; that is, V = (180 V) sin[2π(75 Hz)t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.
The maximum value of the induced magnetic field (Bmax) at a distance r is R from the center of the circular plates is approximately 1.028 × 10^(-7) Tesla.
To find the maximum value of the induced magnetic field (Bmax) at a distance r = R from the center of the circular plates, we can use the formula for the magnetic field generated by a circular loop of current.
The induced magnetic field at a distance r from the center of the circular plates is by:
[tex]B = (μ₀ / 2) * (I / R)[/tex]
where:
B is the magnetic field,
μ₀ is the permeability of free space (approximately [tex]4π × 10^(-7) T·m/A),[/tex]
I is the current flowing through the loop,
and R is the radius of the circular plates.
In this case, the current flowing through the circular plates is by the rate of change of electric charge on the plates with respect to time.
We can calculate the current by differentiating the potential difference equation with respect to time:
[tex]V = (180 V) sin[2π(75 Hz)t][/tex]
Taking the derivative with respect to time:
[tex]dV/dt = (180 V) * (2π(75 Hz)) * cos[2π(75 Hz)t][/tex]
The current (I) can be calculated as the derivative of charge (Q) with respect to time:
[tex]I = dQ/dt[/tex]
Since the charge on the capacitor plates is related to the potential difference by Q = CV, where C is the capacitance, we can write:
[tex]I = C * (dV/dt)[/tex]
The capacitance of a parallel-plate capacitor is by:
[tex]C = (ε₀ * A) / d[/tex]
where:
ε₀ is the permittivity of free space (approximately 8.85 × 10^(-12) F/m),
A is the area of the plates,
and d is the plate separation.
The area of a circular plate is by A = πR².
Plugging these values into the equations:
[tex]C = (8.85 × 10^(-12) F/m) * π * (39 mm)^2 / (3.9 mm) = 1.1307 × 10^(-9) F[/tex]
Now, we can calculate the current:
[tex]I = (1.1307 × 10^(-9) F) * (dV/dt)[/tex]
To find Bmax at r = R, we need to find the current when t = 0. At this instant, the potential difference is at its maximum value (180 V), so the current is also at its maximum:
Imax = [tex](1.1307 × 10^(-9) F) * (180 V) * (2π(75 Hz)) * cos(0) = 2.015 × 10^(-5) A[/tex]
Finally, we can calculate Bmax using the formula for the magnetic field:
Bmax = (μ₀ / 2) * (Imax / R)
Plugging in the values:
Bmax =[tex](4π × 10^(-7) T·m/A / 2) * (2.015 × 10^(-5) A / 39 mm) = 1.028 × 10^(-7) T[/tex]
Therefore, the maximum value of the induced magnetic field (Bmax) at a distance r = R from the center of the circular plates is approximately [tex]1.028 × 10^(-7)[/tex]Tesla.
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When ultraviolet light with wavelength of 300.0 nm falls on certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.60 eV. Find the work function (binding energy) of the metal (in eV).
The work function of the metal is 4.07 eV.
Wavelength of ultraviolet light = 300.0 nm = 3 × 10−7 m
Maximum kinetic energy of photoelectrons = 1.60 eV
Planck's constant = 6.626 × 10−34 J⋅s
Speed of light = 3 × 108 m/s
The energy of the ultraviolet photon is:
E = hν = h / λ = (6.626 × 10−34 J⋅s) / (3 × 10−7 m) = 2.21 × 10−19 J
The work function of the metal is the energy required to remove an electron from the surface of the metal.
It is equal to the difference between the energy of the ultraviolet photon and the maximum kinetic energy of the photoelectrons:
W = E - KE = 2.21 × 10−19 J - 1.60 eV = 4.07 eV
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MY NOTES Three forces acting on an object are given by --1,51 +6.30), F - (4.951 - 1.43), and is - (-40) N. The object experiences an acceleration of magnitude 3.80 m/s2 (a) What is the direction of the acceleration? X Note that the direction of the acceleration is the same as the direction of the net force. (counterclockwise from the +x-axis) (b) What is the mass of the object? kg (cy if the object is initially otrest, what is its speed after 16.03? mus (a) What are the velocity components of the object alter 16.07(Let the velocity be denoted by v 1)) ms Three forces acting on an object are given by 7₁-(-1.51+6.30) N, ₂-(4.951-14) N, and 7-(-441) N. The object experiences an acceleration of magnitude 3.80 m/s². (a) What is the direction of the acceleration? 5.9 x Note that the direction of the acceleration is the same as the direction of the net force." (counterclockwise from the x-axis) (b) What is the mass of the object? kg (c) If the object is initially at rest, what is its speed after 16.0 57 m/s (d) What are the velocity components of the object after 16.0 s? (Let the velocity be denoted by V) 1) mys Need Help? Read Watch
Given,
Three forces acting on an object are given by 7₁-(-1.51+6.30) N, ₂-(4.951-14) N, and 7-(-441) N.
The object experiences an acceleration of magnitude 3.80 m/s².
(a) What is the direction of the acceleration?The net force can be calculated as,
Fnet = F1 + F2 + F3
Fnet = 7 - 1.51 + 6.30 - 4.951 + 1.43 - (-40)N
=> Fnet = 7.87 N
The direction of the net force is counterclockwise from the +x-axis as the force F3 points in the downward direction.
The direction of acceleration will also be in the same direction as the net force.
Therefore, the direction of acceleration is counterclockwise from the +x-axis.
(b) What is the mass of the object?The mass of the object can be calculated as,
m = F / am = Fnet / am
= 7.87 / 3.80m
= 2.07 kg
(c) If the object is initially at rest, what is its speed after 16.0 s?The velocity of the object after 16.0 seconds can be calculated as
v = u + at
u = 0 as the object is at rest
v = at
v = 3.80 x 16v = 60.8 m/s
d) What are the velocity components of the object after 16.0 s?(Let the velocity be denoted by V)
The velocity components of the object can be calculated as,
V = (vx, vy)
Vx can be calculated as, Vx = v × cosθ
Vx = 60.8 × cos5.9°
Vx = 60.73 m/s
Vy can be calculated as, Vy = v × sinθ
Vy = 60.8 × sin5.9°
Vy = 5.58 m/s
Therefore, the velocity components of the object after 16.0 seconds are (60.73 m/s, 5.58 m/s).
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Exercise 2: Mass and Acceleration and 125. 126.4 1261 .3 122.9 wooo Table 4-2: Mass and acceleration for large airtrack glider. acceleration total mass (kg) (m/s) 1/mass (kg') O О 128. Smist 20 125.30 125.5 d 5 4th 113.0 120.0 117.8 121.0 1.9 20 30 30 40 Чо SO 50 60 21.0 misal 118.Oma 117.6ml 115.33 3.3 6th 116.0 117.0 6 115.0 113.2 Attach graph with slope calculation and equation of line clearly written on graph. 2.8 20.7 What does the slope of this line represent? How does the value compare to the measured value (i.e show percent error calculation)? Is the acceleration inversely proportional to the mass? How do you know?
The slope of the line represents the acceleration, and the percent error can be calculated by comparing the measured and theoretical values. The graph helps determine if the acceleration is inversely proportional to the mass.
The slope of a line in a graph represents the rate of change between the variables plotted on the x-axis and y-axis. In this case, the x-axis represents the total mass (kg) and the y-axis represents the acceleration (m/s^2). Therefore, the slope of the line indicates how the acceleration changes with respect to the mass.
To calculate the percent error, the measured value of the slope can be compared to the value obtained from the graph. The percent error can be calculated using the formula:
Percent Error = ((Measured Value - Theoretical Value) / Theoretical Value) * 100
By substituting the measured and theoretical values of the slope into the formula, we can determine the percent error. This calculation helps us assess the accuracy of the measurements and determine the level of deviation between the measured and expected values.
Furthermore, by examining the graph, we can determine whether the acceleration is inversely proportional to the mass. If the graph shows a negative correlation, with a decreasing trend in acceleration as mass increases, then it suggests an inverse relationship. On the other hand, if the graph shows a positive correlation, with an increasing trend in acceleration as mass increases, it indicates a direct relationship. The visual representation of the data in the graph allows us to observe the relationship between acceleration and mass more effectively.
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The two wires shown in Figure P19.38 carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at the following locations.
The net magnetic field at this location will be zero.By plugging in the given values (I = 5.00 A, r = 10.0 cm = 0.1 m), we can calculate the magnitude of the net magnetic field at the specified locations.
To find the net magnetic field at a specific location, we can use the right-hand rule for magnetic fields generated by currents.
At a point equidistant from the two wires, the magnetic fields generated by the two currents will cancel each other out. Therefore, the net magnetic field at this location will be zero.
If the location is closer to one wire than the other, the magnetic field generated by the closer wire will dominate. The direction of the net magnetic field will depend on the direction of the current in that wire.
To determine the magnitude of the net magnetic field, we can use the formula for the magnetic field due to a long, straight wire:
B = (μ0 * I) / (2 * π * r),
where B is the magnetic field, μ0 is the permeability of free space (4π x 10^-7 T·m/A), I is the current, and r is the distance from the wire.
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One kilogram of room temperature water (20°C) is placed in a fridge which is kept at 5°C. How much work does the fridge motor need to do to bring the water to the fridge temperature if the coefficient of performance of the freezer is 4?
Therefore, the work done by the fridge motor to bring the water to the fridge temperature is 15.68 J.
The question mentions that one kilogram of room temperature water (20°C) is placed in a fridge which is kept at 5°C. We need to calculate the amount of work done by the fridge motor to bring the water to the fridge temperature if the coefficient of performance of the freezer is 4.
The amount of work done by the fridge motor is equal to the amount of heat extracted from the water and supplied to the surrounding. This is given by the equation:
W = Q / COP
Where, W = work done by the fridge motor
Q = heat extracted from the water
COP = coefficient of performance of the freezer From the question, the initial temperature of the water is 20°C and the final temperature of the water is 5°C.
Hence, the change in temperature is ΔT = 20°C - 5°C
= 15°C.
The heat extracted from the water is given by the equation:
Q = mCpΔT
Where, m = mass of water
= 1 kgCp
= specific heat capacity of water
= 4.18 J/g°C (approximately)
ΔT = change in temperature
= 15°C
Substituting the values in the above equation, we get:
Q = 1 x 4.18 x 15
= 62.7 J
The coefficient of performance (COP) of the freezer is given as 4. Therefore, substituting the values in the equation
W = Q / COP,
we get:W = 62.7 / 4
= 15.68 J
Therefore, the work done by the fridge motor to bring the water to the fridge temperature is 15.68 J.
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A 100 kg rock is sitting on the ground. A 30.0 kg hyena is
standing on top of it. If the coefficient of friction between the
rock and the ground is 1.963, determine the maximum amount of
friction
A 100 kg rock is sitting on the ground. A 30.0 kg hyena is standing on top of it. If the coefficient of friction between the rock and the ground is 1.963, then the maximum amount of friction is 2504 N.
Given data :
Mass of rock (m1) = 100 kg
Mass of hyena (m2) = 30 kg
Coefficient of friction (μ) = 1.963
The formula to calculate the friction is given as follows : F = μR
where,
F = force of friction
μ = coefficient of friction
R = normal reaction
The normal reaction (R) is equal to the weight of the rock and the hyena which is given as :
R = (m1 + m2) g
where g = acceleration due to gravity (9.8 m/s²)
Putting the given values in the formula :
R = (100 + 30) × 9.8 = 1274 N
To calculate the maximum amount of friction, we multiply the coefficient of friction with the normal reaction :
Fmax = μ R = 1.963 × 1274 ≈ 2504 N
Therefore, the maximum amount of friction is 2504 N.
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Part B What is the current through the 3.00 2 resistor? | ΑΣφ I = A Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part C What is the current through the 6.00 2 resistor? V] ΑΣφ ? I = A Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part D What is the current through the 12.00 resistor? | ΑΣΦ I = A < 1 of 1 Submit Request Answer E = 60.0 V, r = 0 + Part E 3.00 12 12.0 12 Ω What is the current through the 4.00 resistor? ХМУ | ΑΣΦ 6.00 12 4.00 12 I = А
We are given a circuit with resistors of different values and are asked to determine the currents passing through each resistor.
Specifically, we need to find the current through a 3.00 Ω resistor, a 6.00 Ω resistor, a 12.00 Ω resistor, and a 4.00 Ω resistor. The previous answers were incorrect, and we have four attempts remaining to find the correct values.
To find the currents through the resistors, we need to apply Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). Let's go through each resistor individually:
Part B: For the 3.00 Ω resistor, we need to know the voltage across it in order to calculate the current. Unfortunately, the voltage information is missing, so we cannot determine the current at this point.
Part C: Similarly, for the 6.00 Ω resistor, we require the voltage across it to find the current. Since the voltage information is not provided, we cannot calculate the current through this resistor.
Part D: The current through the 12.00 Ω resistor can be determined if we have the voltage across it. However, the given information only mentions the resistance value, so we cannot find the current for this resistor.
Part E: Finally, we are given the necessary information for the 4.00 Ω resistor. We have the voltage (E = 60.0 V) and the resistance (R = 4.00 Ω). Applying Ohm's Law, the current (I) through the resistor is calculated as I = E/R = 60.0 V / 4.00 Ω = 15.0 A.
In summary, we were able to find the current through the 4.00 Ω resistor, which is 15.0 A. However, the currents through the 3.00 Ω, 6.00 Ω, and 12.00 Ω resistors cannot be determined with the given information.
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EM radiation has an average intensity of 1700 W/m2. Which of the following statements about the E or B fields in this radiation is correct? Erms = 800.2 N/C Bmax = 4.42 x 10-6 T Brms = 2.29 x 10-6 T Emax = 1500.0 N/C At a certain place on the surface of the earth, the sunlight has an intensity of about 1.8 x 103 W/m². What is the total electromagnetic energy from this sunlight in 5.5 m³ of space? (Give your answer in joules but don't include the units.) Click Submit to complete this assessment. Question 12 of
The correct statement about the E or B fields in radiation is that Erms = 800.2 N/C.
EM (electromagnetic) radiation has an average intensity of 1700 W/m². As a result, the electrical field (Erms) is related to the average intensity through the equation E = cB, where E is the electric field, B is the magnetic field, and c is the speed of light.
Erms is related to the average intensity I (in W/m²) through the formula Erms = sqrt(2 I / c ε) which is approximately equal to 800.2 N/C.
For a 5.5 m³ space on the earth's surface, the total electromagnetic energy from sunlight with an intensity of 1.8 x 103 W/m² is 9.9 x 106 J.
The formula for calculating the energy is E = I × A × t, where E is the energy, I is the intensity, A is the area, and t is the time.
Here, the area is 5.5 m³ and the time is 1 second, giving an energy of 9.9 x 106 J.
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6) Find the buoyant force on a 0.1 m3 block of wood with density 700 kg/m3 floating in a freshwater lake. (5 pts)
The buoyant force on the 0.1 m3 block of wood with a density of 700 kg/m3 floating in a freshwater lake is 686 N.
Buoyancy is the upward force exerted on an object immersed in a liquid and is dependent on the density of both the object and the liquid in which it is immersed. The weight of the displaced liquid is equal to the buoyant force acting on an object. In this case, the volume of the block of wood is 0.1 m3 and its density is 700 kg/m3. According to Archimedes' principle, the weight of the displaced water is equal to the buoyant force. Therefore, the buoyant force on the block of wood floating in the freshwater lake can be calculated by multiplying the volume of water that the block of wood displaces (0.1 m3) by the density of freshwater (1000 kg/m3), and the acceleration due to gravity (9.81 m/s2) as follows:
Buoyant force = Volume of displaced water x Density of freshwater x Acceleration due to gravity
= 0.1 m3 x 1000 kg/m3 x 9.81 m/s2
= 981 N
However, since the density of the block of wood is less than the density of freshwater, the weight of the block of wood is less than the weight of the displaced water. As a result, the buoyant force acting on the block of wood is the difference between the weight of the displaced water and the weight of the block of wood, which can be calculated as follows:
Buoyant force = Weight of displaced water -
Weight of block of wood
= [Volume of displaced water x Density of freshwater x Acceleration due to gravity] - [Volume of block x Density of block x Acceleration due to gravity]
= [0.1 m3 x 1000 kg/m3 x 9.81 m/s2] - [0.1 m3 x 700 kg/m3 x 9.81 m/s2]
= 686 N
Therefore, the buoyant force acting on the 0.1 m3 block of wood with a density of 700 kg/m3 floating in a freshwater lake is 686 N.
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Bevases of alcohol at room temperature and water that is colder than room temperature are med together in an alted container Select all of the statements that are correct. A The entropies of the water and alcohol each remain unchanged The entropies of the water and alcohol each change, but the sum of their entropies is unchanged The total entropy of the water and alcohol increases The total entropy of the water and cohol decreases E The entropy of the surroundings increases
Bevases of alcohol at room temperature and water that is colder than room temperature are med together in an alted container. The correct statement in this case is B that is the entropies of the water and alcohol each change, but the sum of their entropies is unchanged.
When the warmer alcohol and colder water are mixed together, heat transfer occurs between the two substances. As a result, their temperatures start to equilibrate, and there is an increase in the entropy of the system (water + alcohol). However, the sum of the entropies of the water and alcohol remains unchanged. This is because the increase in entropy of the water is balanced by the decrease in entropy of the alcohol, as they approach a common temperature.
The other statements are incorrect:
A) The entropies of the water and alcohol each remain unchanged - The entropy of the substances changes during the mixing process.
C) The total entropy of the water and alcohol increases - This statement is partially correct. The total entropy of the system (water + alcohol) increases, but the individual entropies of water and alcohol change in opposite directions.
D) The total entropy of the water and alcohol decreases - This statement is incorrect. The total entropy of the system increases, as mentioned above.
E) The entropy of the surroundings increases - This statement is not directly related to the mixing of water and alcohol in an insulated container. The entropy of the surroundings may change in some cases, but it is not directly mentioned in the given scenario.
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A beam of light strikes the surface of glass (n = 1.46) at an angle of 60° with respect to the normal. Find the angle of refraction inside the glass. Take the index of refraction of air n₁ = 1.
The angle of refraction is 53.13°.
Here are the given:
* Angle of incidence: 60°
* Index of refraction of air: n₁ = 1
* Index of refraction of glass: n₂ = 1.46
To find the angle of refraction, we can use the following formula:
sin(θ₂) = n₁ sin(θ₁)
where:
* θ₂ is the angle of refraction
* θ₁ is the angle of incidence
* n₁ is the index of refraction of the first medium (air)
* n₂ is the index of refraction of the second medium (glass)
Plugging in the known values, we get:
sin(θ₂) = 1 * sin(60°) = 0.866
θ₂ = sin⁻¹(0.866) = 53.13°
Therefore, the angle of refraction is 53.13°.
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3. Coulomb's Law refers exclusively to point charges. a. Real b. False
The statement that claims that the Coulomb's Law refers exclusively to point charges is b. False
Coulomb's Law is not limited to point charges; it applies to any charged objects, whether they are point charges or have finite sizes and distributions of charge.
Coulomb's Law states that the magnitude of the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's Law is described by the equation F = k * (q1 * q2) / r^2, where F represents the electrostatic force between two charged objects, k is the electrostatic constant, q1 and q2 denote the charges of the objects, and r signifies the distance separating them.
This law is a fundamental principle in electrostatics and is applicable to a wide range of scenarios involving charged objects, not just point charges.
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A thin film of cooking oil (n = 1.44) is spread on a puddle of water (n = 1.35). What is the minimum thickness Dmin of the oil that will strongly reflect blue light having a wavelength in air of 476 n
The minimum thickness of the cooking oil film that will strongly reflect blue light with a wavelength of 476 nm is approximately 165.3 nm.
To find the minimum thickness Dmin we need to consider the interference of light waves reflected from the top and bottom surfaces of the film.
The refractive indices of the oil and water are given as 1.44 and 1.35, respectively.
When light waves reflect from the top and bottom surfaces of the thin film, interference occurs. For constructive interference (strong reflection), the path length difference between the waves must be an integer multiple of the wavelength.
In this case, the path length difference can be calculated as follows:
2 * n * Dmin = m * λ
where n is the refractive index of the film (cooking oil), Dmin is the minimum thickness of the film, m is an integer representing the order of the interference, and λ is the wavelength of the light in the film.
Since we are interested in the minimum thickness, we can assume m = 1 to find the first-order interference. Therefore:
2 * 1.44 * Dmin = 1 * λ
Substituting the values:
2.88 * Dmin = 476 nm
Dmin = (476 nm) / 2.88
Dmin ≈ 165.3 nm
Therefore, the minimum thickness of the cooking oil film that will strongly reflect blue light with a wavelength of 476 nm is approximately 165.3 nm.
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establish the expansion of a plane wave in terms of an infinite
number spherical waves.
A plane wave can be expanded in terms of an infinite number of spherical waves using a technique called the multipole expansion. The multipole expansion is a mathematical representation that breaks down a complex wave into simpler components.
The expansion begins by considering a plane wave propagating in a specific direction, such as the z-direction. The plane wave can be expressed as:
E_plane(x, y, z) = E0 * exp(i * k * z)
where E0 represents the amplitude of the wave, k is the wave vector, and i is the imaginary unit.
To expand this plane wave into spherical waves, we use the fact that spherical waves can be described as a superposition of plane waves with different directions.
These plane waves have wave vectors that lie along the radial direction in spherical coordinates.
Using spherical coordinates (r, θ, φ), the expansion of the plane wave into spherical waves can be written as:
E_plane(x, y, z) = Σ An * jn(k * r) * Yn,m(θ, φ)
Here, An represents the expansion coefficients, jn is the spherical Bessel function of order n, and Yn,m represents the spherical harmonics.
The sum extends over all possible values of n and m, which results in an infinite series of terms representing spherical waves with different orders and directions.
Each term represents a specific spherical wave with a particular amplitude (given by An), radial dependence (jn(k * r)), and angular dependence (Yn,m(θ, φ)).
The multipole expansion provides a way to describe the plane wave in terms of an infinite number of spherical waves, accounting for the complexity and spatial variation of the original wave.
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Electron density of states inside quantum wires. The electron energy dispersion in an infinite potential barrier quantum wire can be expressed as 2 n E (kr, l, n) = nak? 12,2 2m* 2m* [(+)*+ ()] + where l, n can take integer values 1, 2,.. Derive an expression for the electron density of states and plot this expression for Ly = Ly = 50 Å.
The expression for electron density of states and its plot for Ly = Ly = 50 Å is given below.
Explanation:
To derive an expression for the electron density of states in a quantum wire, we start with the given energy dispersion relation:
E(kr, l, n) = (ħ²k²)/(2m*) + (π²ħ²n²)/(2m*Ly²)
where ħ is the reduced Planck's constant,
k is the wave vector,
m* is the effective mass of the electron,
Ly is the length of the wire in the y-direction,
l is the quantum number related to the quantized transverse modes,
n is the quantum number related to the quantized longitudinal modes.
The electron density of states (DOS) is obtained by calculating the number of allowed states within a given energy range.
In a 1D system, the number of allowed states per unit length in the k-space is given by:
dN(k) = (LxLy)/(2π) * dk
where Lx is the length of the wire in the x-direction.
To find the density of states in energy space, we use the relation:
dN(E) = dN(k) * dk/dE
To calculate dk/dE, we differentiate the energy dispersion relation with respect to k:
dE(k)/dk = (ħ²k)/(m*)
Rearranging the above equation, we get:
dk = (m*/ħ²k) * dE(k)
Substituting this value into the expression for dN(E), we have:
dN(E) = (LxLy)/(2π) * (m*/ħ²k) * dE(k)
Now, we need to express the wave vector k in terms of energy E.
Solving the energy dispersion relation for k, we have:
k(E) = [(2m*/ħ²)(E - (π²ħ²n²)/(2m*Ly²))]^(1/2)
Substituting this value back into the expression for dN(E), we get:
dN(E) = (LxLy)/(2π) * [(m*/ħ²) / k(E)] * dE(k)
Substituting the value of k(E) in terms of E, we have:
dN(E) = (LxLy)/(2π) * [(m*/ħ²) / [(2m*/ħ²)(E - (π²ħ²n²)/(2m*Ly²))]^(1/2)] * dE
Simplifying the expression:
dN(E) = [(LxLy)/(2πħ²)] * [(2m*)^(1/2)] * [(E - (π²ħ²n²)/(2m*Ly²))^(-1/2)] * dE
Now, to obtain the total density of states (DOS), we integrate the above expression over the energy range.
Considering the limits of integration as E1 and E2, we have:
DOS(E1 to E2) = ∫[E1 to E2] dN(E)
DOS(E1 to E2) = ∫[E1 to E2] [(LxLy)/(2πħ²)] * [(2m*)^(1/2)] * [(E - (π²ħ²n²)/(2m*Ly²))^(-1/2)] * dE
Simplifying and solving the integral, we get:
DOS(E1 to E2) = (LxLy)/(πħ²) * [(2m*)^(1/2)] * [(E2 - E1 + (π²ħ²n²)/(2mLy²))^(1/2) - (E1 - (π²ħ²n²)/(2mLy²))^(1/2)]
To plot the expression for the electron density of states, we substitute the given values of Ly and calculate DOS(E) for the desired energy range (E1 to E2), and plot it against energy E.
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What is the best possible coefficient of performance COPret for a refrigerator that cools an environment at -13.0°C and exhausts heat to another environment at 39.0°C? COPrel= How much work W would this ideal refrigerator do to transfer 3.125 x 10 J of heat from the cold environment? W = What would be the cost of doing this work if it costs 10.5¢ per 3.60 × 106 J (a kilowatt-hour)? cost of heat transfer: How many joules of heat Qu would be transferred into the warm environment?
The best possible coefficient of performance (COPret) for the given temperatures is approximately 5.0. The work done by the refrigerator is calculated to be 6.25 x 10 J. The cost of performing this work is approximately 0.0182¢. Finally, the amount of heat transferred into the warm environment is determined to be 9.375 x 10.
The coefficient of performance (COP) of a refrigerator is a measure of its efficiency and is defined as the ratio of the amount of heat transferred from the cold environment to the work done by the refrigerator. For an ideal refrigerator, the COP can be determined using the formula:
COPret = Qc / W
where Qc is the amount of heat transferred from the cold environment and W is the work done by the refrigerator.
To find the best possible COPret for the given temperatures, we need to use the Carnot refrigerator model, which assumes that the refrigerator operates in a reversible cycle. The Carnot COP (COPrel) can be calculated using the formula:
COPrel = Th / (Th - Tc)
where Th is the absolute temperature of the hot environment and Tc is the absolute temperature of the cold environment.
Converting the given temperatures to Kelvin, we have:
Th = 39.0°C + 273.15 = 312.15 K
Tc = -13.0°C + 273.15 = 260.15 K
Substituting these values into the equation, we can calculate the COPrel:
COPrel = 312.15 K / (312.15 K - 260.15 K) ≈ 5.0
Now, we can use the COPrel value to determine the work done by the refrigerator. Rearranging the COPret formula, we have:
W = Qc / COPret
Given that Qc = 3.125 x 10 J, we can calculate the work done:
W = (3.125 x 10 J) / 5.0 = 6.25 x 10 J
Next, we can calculate the cost of doing this work, considering the given cost of 10.5¢ per 3.60 × 10^6 J (a kilowatt-hour). First, we convert the work from joules to kilowatt-hours:
W_kWh = (6.25 x 10 J) / (3.60 × 10^6 J/kWh) ≈ 0.0017361 kWh
To calculate the cost, we use the conversion rate:
Cost = (0.0017361 kWh) × (10.5¢ / 1 kWh) ≈ 0.01823¢ ≈ 0.0182¢
Finally, we need to determine the amount of heat transferred into the warm environment (Qw). For an ideal refrigerator, the total heat transferred is the sum of the heat transferred to the cold environment and the work done:
Qw = Qc + W = (3.125 x 10 J) + (6.25 x 10 J) = 9.375 x 10 J
In summary, the best possible coefficient of performance (COPret) for the given temperatures is approximately 5.0. The work done by the refrigerator is calculated to be 6.25 x 10 J. The cost of performing this work is approximately 0.0182¢. Finally, the amount of heat transferred into the warm environment is determined to be 9.375 x 10.
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An object is 28 cm in front of a convex mirror with a focal length of -21 cm Part A Use ray tracing to determine the position of the image. Express your answer to two significant figures
The position of the image is 12 cm.
To determine the position of the image formed by a convex mirror using ray tracing, we can follow these steps:
Draw the incident ray: Draw a ray from the top of the object parallel to the principal axis. After reflection, this ray will appear to originate from the focal point.
Draw the central ray: Draw a ray from the top of the object that passes through the center of curvature. This ray will reflect back along the same path.
Locate the reflected rays: Locate the intersection point of the reflected rays. This point represents the position of the image.
In this case, the object distance (u) is given as 28 cm (positive because it is in front of the convex mirror), and the focal length (f) is -21 cm. Since the focal length is negative for a convex mirror, we consider it as -21 cm.
Using the ray tracing method, we can determine the position of the image:
Draw the incident ray: Draw a ray from the top of the object parallel to the principal axis. After reflection, this ray appears to come from the focal point (F).
Draw the central ray: Draw a ray from the top of the object through the center of curvature (C). This ray reflects back along the same path.
Locate the reflected rays: The reflected rays will appear to converge at a point behind the mirror. The point where they intersect is the position of the image.
The image formed by a convex mirror is always virtual, upright, and reduced in size.
Using the ray tracing method, we find that the reflected rays converge at a point behind the mirror. This point represents the position of the image. In this case, the position of the image is approximately 12 cm behind the convex mirror.
Therefore, the position of the image is approximately 12 cm.
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Prove the following theorem, known as Bleakney's theorem: If a (nonrelativistic) ion of mass M and initial velocity zero proceeds along some trajectory in given electric and magnetic fields E and B, then an ion of mass kM and the same charge will proceed along the same trajectory in electric and magnetic fields E/k and B. (Hint: Try changing the time scale in the equation of motion for the second ion.)
This can be proven by changing the time scale in the equation of motion for the second ion.M(d²r/dt²) = q(E + v × B) this expression can be used.
Bleakney's theorem states that if a nonrelativistic ion of mass M and initial velocity zero moves along a trajectory in given electric and magnetic fields E and B, then an ion of mass kM and the same charge will follow the same trajectory in electric and magnetic fields E/k and B.
To understand the proof, let's consider the equation of motion for a charged particle in electric and magnetic fields:
M(d²r/dt²) = q(E + v × B)
Where M is the mass of the ion, q is its charge, r is the position vector, t is time, E is the electric field, B is the magnetic field, and v is the velocity vector.
Now, let's introduce a new time scale τ = kt. By substituting this into the equation of motion, we have:
M(d²r/d(kt)²) = q(E + (dr/d(kt)) × B)
Differentiating both sides with respect to t, we get:
M/k²(d²r/dt²) = q(E + (1/k)(dr/dt) × B)
Since the second ion has a mass of kM, we can rewrite the equation as:
(kM)(d²r/dt²) = (q/k)(E + (1/k)(dr/dt) × B)
This equation indicates that the ion of mass kM will experience an effective electric field of E/k and an effective magnetic field of B when moving along the same trajectory. Therefore, the ion of mass kM will indeed follow the same path as the ion of mass M in the original fields E and B, as stated by Bleakney's theorem.
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A 74.6-g ice cube floats in the Arctic Sea. The temperature and pressure of the system and surroundings are at 1 atm and 0°C. Calculate ASsys and ASuniv for the melting of the ice cube in liter-atmosphere per Kelvin. (The molar heat of fusion of water is 6.01 kJ/mol.)
To calculate the entropy change of the system (ASsys) and the total entropy change of the universe (ASuniv) for the melting of the ice cube, we need to consider the heat transfer and the change in entropy.
First, let's calculate the heat transfer during the melting process. The heat transferred is given by the product of the mass of the ice cube, the molar heat of fusion of water, and the molar mass of water. The molar mass of water is approximately 18 g/mol.
Next, we can calculate ASsys using the equation ASsys = q / T, where q is the heat transferred and T is the temperature in Kelvin.
To calculate ASuniv, we can use the equation ASuniv = ASsys + ASsurr, where ASsurr is the entropy change of the surroundings. Since the process is happening at constant pressure and temperature, ASsurr is equal to q / T.
By substituting the calculated values into the equations, we can find the values of ASsys and ASuniv for the melting of the ice cube. The units for entropy change are liter-atmosphere per Kelvin.
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What is the magnitude of the potential difference between two points that are \( 1.46 \mathrm{~cm} \) and \( 2.628 \mathrm{~cm} \) from a proton?
The magnitude of the potential difference between the two points is approximately 0.778 volts (or 0.778 V).
To determine the potential difference between two points, we use the equation:
ΔV = V2 - V1
where ΔV is the potential difference, V2 is the potential at the second point, and V1 is the potential at the first point.
Let's calculate the potential at each of the given points using the equation:
V1 = (9 × 10⁹ N·m²/C²) × (1.6 × 10⁻¹⁹ C / 0.0146 m)
V2 = (9 × 10⁹ N·m²/C²) × (1.6 × 10⁻¹⁹ C / 0.02628 m)
Now, let's substitute the values and calculate:
V1 ≈ 0.824 V
V2 ≈ 0.046 V
Finally, we can calculate the potential difference:
ΔV = V2 - V1 ≈ 0.046 V - 0.824 V ≈ -0.778 V
The negative sign indicates that the potential at the second point is lower than the potential at the first point. However, when we consider the magnitude of the potential difference, we ignore the negative sign.
Therefore, the magnitude of the potential difference between the two points is approximately 0.778 volts (or 0.778 V).
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Consider 0.06 moles of a dialomic ideal gas that undergoes the cycle shown on the pV diagram below. The gas has Cy-2.5A and For this problem we wilt be using the First Law of Thermodynamics used in dess 0+ W where energy transferred into the gas is positive and energy transferred out of the gas is negative. Please make sure you enter the proper plus or minus signs on the answers to each part of this protien Cp-3.5 p (atm) B 3.0 2.0 1.0 V (cm) Part A For process B-C, what is the value of W the work done on the gas by the environment on Joules)? 4 ? VAL 480 Submit Preu A A 800 1600 2400 For process B->C, what is the value of W, the work done on the gas by the environment (in Joules)? 15. ΑΣΦΑ 480 Joules Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Check your signs. Part B For process C->A, what is the value of Q, the heat absorbed/released by the gas (in Joules)? 17 ΑΣΦ Joules Request Answer Submit Y Part C For the entire cycle (A-B, BC, CA), what is the total heat absorbed/released by the gas, Ororin Joules)?
Since the area is below the axis, the work done on the gas is negative and the answer is -15 J.
For process, B-C, the work done on the gas by the environment is determined by the area under the curve. As shown on the graph, the area is a trapezoid, so the formula for its area is ½ (b1+b2)h. ½ (2 atm + 1 atm) x (10 cm - 20 cm) = -15 J. Since the area is below the axis, the work done on the gas is negative.
Therefore, the answer is -15 J.
For process, C-A, the heat absorbed/released by the gas is equal to the negative of the heat absorbed/released in process A-B. Thus, Q = -17 J. The negative sign implies that the heat is released by the gas in this process.
For the entire cycle, the net work done is the sum of the work done in all three processes. Therefore, Wnet = Wbc + Wca + Wab = -480 J + 15 J + 465 J = 0. Qnet = ΔU + Wnet, where ΔU = 0 (since the gas returns to its initial state). Therefore, Qnet = 0.
For process B-C, the value of W, the work done on the gas by the environment, is -15 J. For process, C-A, the value of Q, the heat absorbed/released by the gas, is -17 J. For the entire cycle, the net work done is 0 and the net heat absorbed/released by the gas is also 0.
In the pV diagram given, the cycle for a diatomic ideal gas with Cp = 3.5 R and Cy = 2.5 R is shown. The given cycle has three processes: B-C, C-A, and A-B. The objective of this question is to determine the work done on the gas by the environment, W, and the heat absorbed/released by the gas, Q, for each process, as well as the network and heat for the entire cycle. The first law of thermodynamics is used for this purpose:
ΔU = Q - W. For any cycle, ΔU is zero since the system returns to its initial state. Therefore, Q = W. For process, B-C, the work done on the gas by the environment is determined by the area under the curve. The area is a trapezoid, and the work is negative since it is below the axis. For process, C-A, the heat absorbed/released by the gas is equal to the negative of the heat absorbed/released in process A-B. The work done by the gas is equal to the work done on the gas by the environment since the process is the reverse of B-C. The net work done is the sum of the work done in all three processes, and the net heat absorbed/released by the gas is zero since Q = W.
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An object is rotating in a circle with radius 2m centered around the origin. When the object is at location of x = 0 and y = -2, it's linear velocity is given by v = 2i and linear acceleration of q = -3i. which of the following gives the angular velocity and angular acceleration at that instant?
The angular velocity at that instant is 1 rad/s and the angular acceleration is -1.5 rad/s².
To determine the angular velocity and angular acceleration at the instant, we need to convert the linear velocity and linear acceleration into their corresponding angular counterparts.
The linear velocity (v) of an object moving in a circle is related to the angular velocity (ω) by the equation:
v = r * ω
where:
v is the linear velocity,
r is the radius of the circle,
and ω is the angular velocity.
The radius (r) is 2m and the linear velocity (v) is 2i, we can find the angular velocity (ω):
2i = 2m * ω
ω = 1 rad/s
So, the angular velocity at that instant is 1 rad/s.
Similarly, the linear acceleration (a) of an object moving in a circle is related to the angular acceleration (α) by the equation:
a = r * α
where:
a is the linear acceleration,
r is the radius of the circle,
and α is the angular acceleration.
The radius (r) is 2m and the linear acceleration (a) is -3i, we can find the angular acceleration (α):
-3i = 2m * α
α = -1.5 rad/s²
Therefore, the angular velocity at that instant is 1 rad/s and the angular acceleration is -1.5 rad/s².
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An electron has a charge of 1.60x101°C. At what distance from the electron would the magnitude of the electric field be 5.14×10"N/C? What is the direction of the
field there? (5.29x10-1 m, toward the electron)
To calculate the distance from the electron where the magnitude of the electric field is a specific value, we can use Coulomb's law and rearrange the formula to solve for distance.
Coulomb's law states:
E = k * (|q| / r^2)
where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), |q| is the magnitude of the charge, and r is the distance from the charge.
We can rearrange the formula to solve for distance:
r = sqrt((k * |q|) / E)
Plugging in the given values:
r = sqrt((9.0 x 10^9 N m^2/C^2 * 1.60 x 10^(-19) C) / (5.14 x 10^6 N/C))
Simplifying:
r = sqrt((9.0 x 1.60 x 10^(-19) / 5.14 x 10^6) * 10^9 m^2/C^2)
r = sqrt((14.4 x 10^(-19)) / 5.14 x 10^6) * 10^9 m
r = sqrt(2.80 x 10^(-25)) * 10^9 m
r ≈ sqrt(2.80) * 10^(-8) m
r …
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