Explanation:
The given data is as follows.
C = [tex]20 \times 10^{-6} F[/tex]
R = [tex]100 \times 10^{3}[/tex] ohm
[tex]Q_{o} = 100 \times 10^{-6}[/tex] C
Q = [tex]13.5 \times 10^{-6} C[/tex]
Formula to calculate the time is as follows.
[tex]Q_{t} = Q_{o} [e^{\frac{-t}{\tau}][/tex]
[tex]13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}][/tex]
0.135 = [tex]e^{\frac{-t}{2}}[/tex]
[tex]e^{\frac{t}{2}} = \frac{1}{0.135}[/tex]
= 7.407
[tex]\frac{t}{2} = ln (7.407)[/tex]
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.
The required time to discharge the capacitor to a certain charge can be calculated by rearranging the exponential decay formula. The product of resistance and capacitance, known as the RC time constant, is a key value in this calculation. After substituting the values given into the formula, we get the required time.
Explanation:The time it takes for a capacitor to discharge through a resistor can be calculated using the formula for the decay of charge on a capacitive circuit: Q = Q0 * e-t/RC, where Q is the charge at time t, Q0 is the initial charge, R is the resistance and C is the capacitance. To find the time at which the charge will be 13.5 uC, we rearrange the formula to solve for time t: t = -RC * ln(Q/Q0).
Using the values of the problem: R = 100 kΩ = 100,000 Ω, C = 20 uF = 20*10-6 F, Q0 = 100 uC = 100*10-6 C, and Q = 13.5 uC = 13.5*10-6 C, substitute these values into the equation: t = -100,000 * 20*10-6 * ln (13.5/100). Hence, the time it takes for a capacitor to discharge to a specified charge can be calculated using exponential decay formula based on the capacitor's unique RC time constant.
Learn more about Capacitor Discharge here:https://brainly.com/question/33821590
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A lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block?(unit=kg)
Mass of the block is 0.557 kg
Solution:
The quantity of heat (q) liberated during any process is equal to the product of the mass of the block (m), specific heat (C) and the change in temperature (ΔT). Specific heat of lead is 130 J/kg °C.
We have to rearrange the equation to find the mass of the block of lead as,
[tex]\begin{array}{l}\boldsymbol{q}=\boldsymbol{m} \times \boldsymbol{C} \times \boldsymbol{\Delta} \mathbf{T}\\ \\\boldsymbol{m}=\frac{\mathbf{q}}{\boldsymbol{c} \times \mathbf{\Delta} \mathbf{T}}\\ \\\mathbf{m}=\frac{427 \mathbf{J}}{13 \mathbf{0}_{\mathbf{4} *}^{\prime} \mathbf{r} \times \mathbf{s} \mathbf{9} \mathbf{0}^{*} \mathbf{c}}=\mathbf{0} . \mathbf{5} \mathbf{5} \mathbf{7} \mathbf{k} \mathbf{g}\end{array}[/tex]
Thus the mass (m) = 0.577 kg
Answer:
the correct answer is 0.57
Explanation:
Describe what changes you would expect to see in a plot of absorbance versus temperature in an optical melting experiment on a self-complementary duplex.
Answer:
#Check explanation below for a detailed description.
Explanation:
A plot of absorbance vs temperature is used to determine [tex]T_m[/tex] of a duplex DNA.
-The duplex DNA denatures.
-The denaturation creates two single strands which enhances Ultraviolet absorption ( increased temperatures increases absorption rates).
-Higher concentrations of [tex]Sodium \ Chloride[/tex] hampers stability of the DNA.
When light goes from one material into another material having a HIGHER index of refraction,
A) its speed decreases but its wavelength and frequency both increase.
B) its speed, wavelength, and frequency all decrease.
C) its speed increases, its wavelength decreases, and its frequency stays the same.
D) its speed decreases but its frequency and wavelength stay the same.
E) its speed and wavelength decrease, but its frequency stays the same.
Answer:
When light goes from one material into another material having a HIGHER index of refraction, its speed and wavelength decrease, but its frequency stays the same.
The correct option is E
Explanation:
Refraction is the bending of light ray as it crosses the boundary between the two media of different densities, thus causing a change in it's direction.
When light goes from one material of lower refractive to another material of higher refractive index, the speed of light reduces, because from law of refraction
aNg= (speed of light in air)/(speed of light in glass)
Snell's Law
So generally,
n1/n2 = v2/v1
Let n1 be incident index
Let n2 be refracted index
v2 is refracted speed
v1 is incident speed
Since given that, n1<n2
Then, the right side of the equation will always be less than 1 i.e n1/n2
Then, v2=v1• ( n1/n2)
Since n1/n2 is less that one this show that the speed v2 will reduce.
Since we know that the speed decrease,
We also know that speed, wavelength and speed is related by
v=fλ
Since, the speed is directly proportional to wavelength this shows that as the speed reduces the wavelength also reduces where frequency is the constant of proportionality
v∝λ. .
then f is constant of proportional
v=fλ
So as speed reduces, the wavelength reduces.
frequency of a light wave does not depend on the medium, while wavelength and speed do.
its speed and wavelength decrease, but its frequency stays the same.
Then the answer is E
If a trapeze artist rotates 1 each second while sailing through the air, and contracts to reduce her rotational inertia to 0.40 of what it was, how many rotations per second will result?
Explanation:
A trapeze is rotating with 1 rotation per second .
Thus its angular velocity ω = 2π n
here n is the number of rotations per second
Thus ω = 2π b because n = 1 in this case
Suppose the moment of inertia of his is = I
Then angular momentum L₁ = I ω = 2 I π
In the second case , the moment of inertia becomes = 0.4 I
Let his angular velocity is ω₀
Thus angular momentum L₂ = 0.4 I ω₀
Because no external torque is applied , therefore angular momentum will remain constant .
Thus L₁ = L₂
Therefore 2 I π = 0.4 I x 2 n₀ π
here n₀ is the number of rotations per second
n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5
What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 46 rpmrpm (revolutions per minute) if the wheel's diameter is 32 cmcm ?
Acceleration of a speck is 0.77 m/s²
Explanation:
Given of the solution-
Diameter (We can represent as d) = 32 cm
radius, (We can represent as r )= 32/2 = 16 cm = 0.16 m
Angular acceleration,(We can represent as α ) = 46 rpm
[tex]\alpha = 46 * \frac{2\pi }{60}[/tex]
Acceleration, a = ?
We know that the formula for the acceleration is
[tex]a = r * \alpha[/tex]
[tex]a = 0.16 * 46 * \frac{2\pi }{60}\\ \\a = 0.16 * 46 * \frac{2 * 3.14}{60}[/tex]
[tex]a = 0.77m/s^2[/tex]
Therefore, acceleration of a speck is 0.77 m/s²
Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates separated by 1.6 mm and having a potential difference of 148 V. The magnetic field between the plates is 0.42 T. The magnetic field in the mass spectrometer is 1.2 T.
1)(a) Find the speed of the ions entering the mass spectrometer.
m/s
2)(b) Find the difference in the diameters of the orbits of singly ionized 238U and 235U. (The mass of a 235U ion is 3.903 x 10-25 kg.)
mm
Answer:
1) v = 2.20 10⁵ m / s , 2) r = 4.86 10⁻³ m, r = 4.92 10⁻³ m
Explanation:
1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so
[tex]F_{m} = F_{e}[/tex]
q v B = q E
v = E / B
V = E s
E = V / s
v = V / s B
v = 148 / (0.0016 0.42)
v = 2.20 10⁵ m / s
2) when the isotopes enter the spectrometer, we can use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
q v B = m v² / r
r = m v / qB
The mass of uranium 235 is
m = 3.903 10⁻²⁵ kg
The radius of this isotope is
r = 3,903 10⁻²⁵ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.86 10⁻³ m
The mass of the uranium isotope 238 is
m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg
The radius is
r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.92 10⁻³ m
Final answer:
The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.
Explanation:
To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.
Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.
To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.
When there is a large difference in two resistor’s sizes, what useful approximations can be used when considering their series and parallel combinations? (This is a handy thing to know in circuit design!)
Answer:
Series circuit approximation = R1 >> R2 then R2 ≈ 0
Parallel circuit approximation = R1 << R2 then R2 ≈ 0
Explanation:
in a series circuit, if there is a large difference between two resistors then we can omit the resistor which has low resistance because the higher value resistor dominates.
R1 >> R2 then R2 ≈ 0
in a parallel circuit, if there is a large difference between two resistors then we can omit the resistor which has high resistance because the lower value resistor dominates.
R1 << R2 then R2 ≈ 0
A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 0.9 µF capacitor then drops to 2 V. What is the capacitance of the second capacitor?
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
A circular coil with 169 turns has a radius of 2.6 cm. (a) What current through the coil results in a magnetic dipole moment of 3.4 A·m2? I= A (b) What is the maximum torque that the coil will experience in a uniform field of strength 0.03 T? τmax= Nm (c) If the angle between μ and B is 44.9∘, what is the magnitude of the torque on the coil? τ= Nm (d) What is the magnetic potential energy of coil for this orientation?
Answer with Explanation:
We are given that
Number of turns=n=169
Radius of coil=r=2.6 cm=[tex]2.6\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2}m[/tex]
Area of circular coil=[tex]\pi r^2[/tex]
Where [tex]\pi=3.14[/tex]
Using the formula
Area of circular coil=[tex]3.14\times (2.6\times 10^{-2})^2=21.2\times 10^{-4}m^2[/tex]
a.Magnetic dipole moment=[tex]\mu=3.4 Am^2[/tex]
[tex]I=\frac{\mu}{nA}[/tex]
Using the formula
[tex]I=\frac{3.4}{169\times 21.2\times 10^{-4}}[/tex]
[tex]I=9.5 A[/tex]
b.Magnetic field, B=0.03 T
[tex]\tau_{max}=\mu B[/tex]
Substitute the value
[tex]\tau_{max}=3.4\times \times 0.03=0.102 Nm[/tex]
c.[tex]\theta=44.9^{\circ}[/tex]
[tex]\tau=\tau_{max}sin\theta[/tex]
Substitute the values
[tex]\tau=0.102sin44.9=0.07 Nm[/tex]
A thin aluminum rod lies along the x-axis and has current of I = 16.0 A running through it in the +x-direction. The rod is in the presence of a uniform magnetic field, perpendicular to the current. There is a magnetic force per unit length on the rod of 0.113 N/m in the −y-direction.
(a) What is the magnitude of the magnetic field (in mT) in the region through which the current passes?
(b) What is the direction of the magnetic field in the region through which the current passes?
Answer:
a) The magnitude of the magnetic field = 7.1 mT
b) The direction of the magnetic field is the +z direction.
Explanation:
The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by
F = (B)(I)(L) sin θ
F/L = (B)(I) sin θ
For this question,
(F/L) = 0.113 N/m
B = ?
I = 16.0 A
θ = 90°
0.113 = B × 16 × sin 90°
B = 0.113/16 = 0.0071 T = 7.1 mT
b) The direction of the magnetic field will be found using the right hand rule.
The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).
An electric motor is plugged into a standard wall socket and is running at normal speed. Suddenly, some dirt prevents the shaft of the motor from turning quite so rapidly. What happens to the back emf of the motor, and what happens to the current that the motor draws from the wall socket?
a.) The back emf decreases, and the current drawn from the socket decreases.
b.) The back emf increases, and the current drawn from the socket decreases.
c.) The back emf decreases, and the current drawn from the socket increases.
d.) The back emf increases, and the current drawn from the socket increases.
Final answer:
The back emf of the motor decreases due to reduced speed caused by dirt in the shaft, which, in turn, causes the current drawn from the wall socket to increase.
Explanation:
When dirt prevents the shaft of an electric motor from turning rapidly, the back emf produced by the motor decreases. Since the back emf is proportional to the motor's angular velocity, any decrease in speed results in a decrease in back emf. Because the back emf opposes the applied voltage from the wall socket, a decrease in back emf implies that the voltage across the motor's coil will increase, causing the motor to draw a larger current from the socket. Therefore, the current that the motor draws from the wall socket will increase. The correct answer to the question is (c) The back emf decreases, and the current drawn from the socket increases.
A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from turkey.?
What is the velocity of the ball after the kick?
What force did the turkey feel?
Why doesn’t the turkey go flying backward since the ball flies forward? Is momentum conserved?
Answer:
a. 20m/s
b.50N
c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.
Explanation:
a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:
[tex]F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s[/tex]
Hence, the velocity of the ball after the kick is 20m/s
b.The force felt by the turkey:
#Applying Newton's 3rd Law of motion, opposite and equal reaction:
-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.
c. Using the law of momentum conservation:
-Due to ther external forces exerted on the turkey, it remains stationery.
-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.
-Momentum is not conserved due to these external forces.
The velocity of the ball after the kick is 200 m/s. The turkey feels a force of 50N. The turkey doesn't go flying backward because of its greater mass compared to the ball.
Explanation:The velocity of the ball after the kick can be calculated using the equation:
velocity = force / mass * time
Substituting the values, we get:
velocity = 50N / 0.5kg * 0.2s = 200 m/s
The force that the turkey feels can be determined using Newton's third law, which states that for every action, there is an equal and opposite reaction. So, the force the turkey feels will be the same as the force of the kick, which is 50N.
The turkey doesn't go flying backward because the turkey has a greater mass than the ball. According to Newton's second law, the acceleration of an object is inversely proportional to its mass. Since the turkey has a greater mass, it experiences less acceleration compared to the ball. Conservation of momentum is upheld in this situation, as the momentum of the turkey and ball together remains the same before and after the kick.
A point charge of -3.0X10-5 C is placed at the origin of coordinates in vacuum. Find the electric field at the point x= 5.0 m on the x axis. Determine the acceleration of a proton (q=+e, m=1.67X10-27 kg) immersed in an electric field of strength 0.50 kN/C in vacuum. How many times greater is this acceleration than that due to gravity?
a) -10,800 N/C
b) [tex]4.79\cdot 10^{10}m/s^2[/tex]
c) [tex]4.88\cdot 10^9[/tex] times g
Explanation:
a)
The magnitude of the electric field produced by a single-point charge is given by
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge producing the field
r is the distance at which the field is calculated
In this problem:
[tex]Q=-3.0\cdot 10^{-5}C[/tex] is the c harge producing the field
[tex]r=5.0 m[/tex] is the distance at which we want to calculate the field
Substituting,
[tex]E=\frac{(9.0\cdot 10^9)(-3.0\cdot 10^{-5})}{(5.0)^2}=-10,800 N/C[/tex]
where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)
b)
The force experienced by a charged particle in an electric field is given by
[tex]F=qE[/tex]
where
q is the magnitude of the charge
E is the electric field
Moreover, the force on an object can be written as:
[tex]F=ma[/tex]
where
m is the mass
a is the acceleration
Combining the two equations,
[tex]ma=qE\\a=\frac{qE}{m}[/tex]
In this problem:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]E=0.50 kN/C=500 N/C[/tex] is the strength of the electric field
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the mass of the proton
Substituting, we find the acceleration of the proton:
[tex]a=\frac{(1.6\cdot 10^{-19})(500)}{(1.67\cdot 10^{-27})}=4.79\cdot 10^{10}m/s^2[/tex]
c)
The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by
[tex]g=9.81 m/s^2[/tex]
On the other hand, the acceleration of the proton in this problem is:
[tex]a=4.79\cdot 10^{10} m/s^2[/tex]
To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:
[tex]\frac{a}{g}=\frac{4.79\cdot 10^{10}}{9.81}=4.88\cdot 10^9[/tex]
So, it is [tex]4.88\cdot 10^9[/tex] times greater than g.
Final answer:
The electric field at the given point is -1.08×10 N/C, which points towards the origin. The acceleration of a proton in a 0.50 kN/C electric field is 4.79×10ⁱ³ m/s². This acceleration is approximately 4.88×10ⁱ¹ times greater than the acceleration due to gravity.
Explanation:
To calculate the electric field at a point due to a point charge, we use Coulomb's law and the expression for the electric field, E = kQ/r², where k is Coulomb's constant (8.99×10⁹ N⋅m²/C²), Q is the charge, and r is the distance from the charge. Placing a charge of -3.0×10⁻⁵ C at the origin, the electric field at the point x= 5.0 m on the x-axis is calculated as follows:
E = kQ/r²
= (8.99×10⁹)(-3.0×10⁻⁵) / (5.0)²
= -1.08×10 N/C
The electric field points towards the negative charge, which is towards the origin.
To determine the acceleration of a proton in an electric field, we use the formula F = qE, where F is the force, q is the charge of the proton (1.60×10⁻⁵ C), and E is the electric field strength. The acceleration a is then found using Newton's second law, F = ma:
F = qE
= (1.60×10⁻⁵)(0.50×10)
= 8.00×10⁻⁴ N
a = F/m
= 8.00×10⁻⁴ / 1.67×10⁻⁷
= 4.79×10ⁱ³ m/s²
To find how many times greater this is compared to the acceleration due to gravity, we simply divide the proton's acceleration by the acceleration due to gravity (9.81 m/s²).
Number of times greater = a / g
= 4.79×10ⁱ³ / 9.81
=
≈ 4.88×10ⁱ¹ times
A beam of electrons is accelerated from rest through a potential difference of 0.200 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at ± 13.6 ∘ from the original direction of the beam.
Do we need to use relativity formulas? Select the correct answer and explanation.
a. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ−1>>1 which means that we do not have to use special relativity.
b. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ<<1 which means that we have to use special relativity.
c. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ>>1 which means that we have to use special relativity.
d. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is 0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ−1<<1 which means that we do not have to use special relativity.
Part B
How wide is the slit?
Option d is correct; relativity formulas are not needed because the Lorentz factor (γ-1) is much less than 1, indicating that the electron's velocity is non-relativistic after being accelerated through a potential difference of 0.200 kV.
Explanation:We need to determine whether we should use relativity formulas for the electrons accelerated through a potential difference. When electrons gain kinetic energy (K) by acceleration through a potential difference (V), the energy they gain can be expressed as Ve=K=(γ-1)mc2. Here, γ stands for Lorentz factor, m is the mass of electron, and c is the speed of light. Given that the potential difference ('V') is 0.200 kV, and the rest mass energy ('Eo') of an electron is 0.511 MeV, we calculate:
γ-1= (0.200 keV) / (0.511 MeV) ≈ 0.3916 × 10-3, which implies that γ is approximately 1. So, γ-1 << 1, which in turn means that the electron's velocity is much less than the speed of light, and the relativistic effect can be neglected. Thus, we do not need to use special relativity.
The correct answer is d.
A charge Q is transfered from an initially uncharged plastic ball to an identical ball 12 cm away. The force of attraction is then 17 mN. How many electrons were transfered from one ball to the other
Answer:
Explanation:
The ball from which electrons are transferred will acquire Q charge and the ball receiving electrons will acquire - Q charge . force of attraction between them
= k Q² / d²
9 x 10⁹ x Q² / .12² = 17 X 10⁻³
Q² = .0272 X 10⁻¹²
Q = .1650 X 10⁻⁶ C
No of electrons = .1650 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .103125 x 10¹³
1.03125 x 10¹²
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?
Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by
[tex]V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }[/tex]
Taking natural log on both sides,
[tex]e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })[/tex]
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex] (1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= [tex]\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }[/tex]
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex]
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant 37.0 N/cm and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 66.0 kg are pushed against the other end, compressing the spring 0.370 m. The sled is then released with zero initial velocity.
What is the sled's speed when the spring is still compressed 0.180 m?
Answer:
2.2 m/s
Explanation:
solution:
To calculate change in stored energy at desired extension
ΔU = 1/2*k*(δx)^2
= 1/2*3700*(0.37^2-0.180^2)
= 201 N.m
use work energy theorem
ΔU = ΔK = 1/2*m*v^2 = 201
= 2.2 m/s
note:
calculation maybe wrong but method is correct.
Explanation:
Below is an attachment containing the solution.
You shoot a 50.3-g pebble straight up with a catapult whose spring constant is 320.0 N/m. The catapult is initially stretched by 0.190 m. How high above the starting point does the pebble fly? Ignore air resistance.
Answer:
The pebble reaches a height of 11.716 m above the starting point.
Explanation:
Given:
Mass of pebble (m) = 50.3 g = 0.0503 kg [1 g = 0.001 kg]
Spring constant (k) = 320.0 N/m
Elongation of catapult (x) = 0.190 m
Height of fly (h) = ?
Air resistance is ignored. So, conservation of energy holds true.
The energy stored in the catapult on stretching it is elastic potential energy. This elastic potential energy is transferred to the pebble in the form of gravitational potential energy. Therefore,
Elastic potential energy = Gravitational potential energy
[tex]\frac{1}{2}kx^2=mgh\\\\h=\frac{kx^2}{2mg}[/tex]
Plug in the given values and solve for 'h'. This gives,
[tex]h=\frac{(320.0\ N/m)(0.190\ m)^2}{2(0.0503\ kg)(9.8\ m/s^2)}\\\\h=\frac{11.552\ Nm}{0.986\ N}\\\\h=11.716\ m[/tex]
Therefore, the pebble reaches a height of 11.716 m above the starting point.
The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is doubled without altering the charge on the plates, what is the new potential difference between the plates? Answer in units of V.
Answer:
Explanation:
capacitance of parallel plate capacitor
c = ε A / d , d is distance between plates , A is surface area , ε is constant
As d becomes two times , Capacitance c = 1/ 2 times ie c / 2
potential V = Q / C
Q is constant , potential
v = Q / c /2
= 2 . Q / C
= 2 V
So potential difference becomes 2 times.
NEW P D = 398 X 2
= 796 V.
g Case 1, a mass M M hangs from a vertical spring having spring constant k , k, and is at rest at its equilibrium height. In Case 2, the same mass has been lifted a distance D D vertically upward. If the potential energy in Case 1 is defined to be zero, what is the potential energy in Case 2
In the scenario described, increasing the height of a mass M in a vertical spring-mass system will result in both gravitational and spring potential energy. Thus, the potential energy in Case 2 is the sum of gravitational potential energy (m*g*D) and the elastic potential energy of the spring (1/2*k*D^2).
Explanation:The situation described in the question deals with principles of Physics, specifically potential energy and its application to the spring-mass system. Consider both cases with the mass M hanging from a spring with a constant k in a vertical orientation. In the first case, the system is at equilibrium and thus the potential energy is defined as zero.
In the second case, the mass M is lifted upward a vertical distance D from the equilibrium position, it now possesses gravitational potential energy in addition to the elastic potential energy of the spring. This total energy is preserved as long as no external force behaves on the system. The gravitational potential energy gained by the mass is equal to m*g*D where m is the mass, g is the acceleration due to gravity and D is the vertical distance lifted.
The potential energy stored in the spring when it is stretched or compressed by a distance 'x' is given by the formula U = 1/2*k*x^2, where 'k' is the spring constant and 'x' is the displacement from the equilibrium position (in this case is D).
So, combining both energies, the total potential energy in Case 2, when the mass has been lifted a vertical distance D is m*g*D + 1/2*k*D^2.
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A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t = 0, the left end of the string has its maximum upward displacement.
(a) What are the frequency and angular frequency of the wave?
(b) What is the wave number of the wave?
Answer:
(a)
[tex]f=20Hz\\And \\w=125.7s^{-1}[/tex]
(b)
[tex]K=3.490m^{-1}[/tex]
Explanation:
We know that the speed of any periodic wave is given by:
v=f×λ
The wave number k is given by:
K=2π/λ
Given data
Amplitude A=2.50mm
Wavelength λ=1.80m
Speed v=36 m/s
For Part (a)
For the wave frequency we plug our values for v and λ.So we get:
v=f×λ
[tex]36.0m/s=(1.80m)f\\f=\frac{36.0m/s}{1.80m}\\ f=20Hz\\[/tex]
And the angular speed we plug our value for f so we get:
[tex]w=2\pi f\\w=2\pi (20)\\w=125.7s^{-1}[/tex]
For Part (b)
For wave number we plug the value for λ.So we get
K=2π/λ
[tex]K=\frac{2\pi }{1.80m}\\ K=3.490m^{-1}[/tex]
The answers for this question are
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Given to us:
Amplitude A= 2.50 mm,
Wavelength λ= 1.80 m,
Velocity V= 36.0 m/sec,
a.) To find out frequency and angular frequency of the wave,we know
[tex]frequency=\dfrac{Velocity}{Wavelength}[/tex]
[tex]f=\dfrac{V}{\lambda}[/tex]
Putting the numerical value,
[tex]f=\dfrac{36}{1.80}\\\\f= 20\ Hz[/tex]
Therefore, the frequency of this wave is 20 Hz.
Also, for angular frequency
[tex]f=2\pi \omega[/tex]
Putting the numerical value,
[tex]\omega=2\pi f \\\\\omega=2\times\pi \times 20\\\\\omega= 125.71428\ s^{-1}[/tex]
Therefore, the angular frequency of this wave is [tex]125.71428\ s^{-1}[/tex].
b.) To find out the wave number of the wave (k),
The wave number for an EM field is equal to 2π divided by the wavelength(λ) in meters.
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
Putting the numerical value,
[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{1.80}\\\\k=3.4920\ m^{-1}[/tex]
Therefore, the wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Hence, for this wave
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
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An object with total mass mtotal = 15.8 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.5 kg moves up and to the left at an angle of θ1 = 18° above the –x axis with a speed of v1 = 27.5 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 23° to the right of the -y axis at a speed of v2 = 21.4 m/s. 1)What is the magnitude of the final momentum of the system (all three pieces)?
Answer:
Explanation:
total mass, M = 15.8 kg
initial velocity, u = 0 m/s
m1 = 4.5 kg
m2 = 5.4 kg
v1 = 27.5 m/s at an angle 18°
v2 = 21.4 m/s at an angle 23°
As there is no external force acts on the system, so the moemntum of the system is conserved.
Momentum before collision = momentum after collision
as the momentum before collision is zero, so the momentum after collision is also zero.
The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which [tex]\mu k[/tex] = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant [tex]\theta = 30^{o}[/tex].
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
[tex]\sum F_{y} = 0[/tex]
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
[tex]\sum F_{x} = 0[/tex]
[tex]T_{2} - F_{k}[/tex] = 0
[tex]T_{2} = \mu_{k} N[/tex]
= [tex]0.4 \times 75 lb[/tex]
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
[tex]\sum F_{x} = 0[/tex]
[tex]T Cos (30^{o}) + T Cos (30^{o}) = T_{2}[/tex]
[tex]2T Cos (30^{o}) = T_{2}[/tex]
or, T = [tex]\frac{T_{2}}{2 Cos (30^{o})}[/tex]
= [tex]\frac{30}{2 Cos (30^{o})}[/tex]
= [tex]\frac{30}{1.732}[/tex]
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
= [tex]17.32 lb \times 6[/tex]
= [tex]103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}[/tex]
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
Answer:
The out put power is 0.188 hp.
Explanation:
Given that,
Weight = 75 lb
Coefficient of friction = 0.4
Rate = 6 ft/s
Suppose, Determine the output of the motor at the instant θ = 30°.
For block,
We need to calculate the force in vertical direction
Using balance equilibrium equation in vertical
[tex]\sum{F_{y}}=0[/tex]
[tex]N-W=0[/tex]
[tex]N=W[/tex]
Put the value into the formula
[tex]N=75\ lb[/tex]
Using balance equilibrium equation in horizontal
[tex]\sum{F_{x}}=0[/tex]
[tex]T_{2}-f_{k}=0[/tex]
[tex]T_{2}=\mu_{k}N[/tex]
Put the value into the formula
[tex]T_{2}=0.4\times75[/tex]
[tex]T_{2}=30\ lb[/tex]
For pulley,
We need to calculate the force
Using balance equilibrium equation in horizontal
[tex]\sum{F_{x}}=0[/tex]
[tex]T\cos\theta+T\cos\theta=T_{2}[/tex]
[tex]2T\cos30=T_{2}[/tex]
[tex]T=\dfrac{T_{2}}{2\cos30}[/tex]
Put the value into the formula
[tex]T=\dfrac{30}{2\times\cos30}[/tex]
[tex]T=17.32\ lb[/tex]
We need to calculate the out put power
Using formula of power
[tex]P=Tv[/tex]
Put the value into the formula
[tex]P=17.32\times6[/tex]
[tex]P=103.92\ lb.ft/s[/tex]
[tex]P=0.188\ hp[/tex]
Hence, The out put power is 0.188 hp.
What are (a) the x component and (b) the y component of a vector in the xy plane if its direction is 259° counterclockwise from the positive direction of the x axis and its magnitude is 5.4 m?
Answer:
|Ax| =1.03 m (directed towards negative x-axis)
|Ay|= 5.30 (directed towards negative y-axis)
Explanation:
Let A is a vector = 5.4 m
θ = 259°
to Find Ax, Ay
Sol:
according the condition it lies in 3rd quadrant
we know that Horizontal Component Ax = A cos θ
Ax = 5.4 Cos 259°
Ax = - 1.03 m
|Ax| =1.03 m (directed towards negative x-axis)
Now Ay = A sin θ
Ay = 5.4 Sin 259°
Ay = -5.30
|Ay|= 5.30 (directed towards negative y-axis)
(a) -1.030m
(b) -5.301m
Explanation:Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;
The x-component ([tex]F_{X}[/tex]) of vector F is given by;
[tex]F_{X}[/tex] = F cos θ ---------------------(i)
And;
The y-component ([tex]F_{Y}[/tex]) of vector F is given by;
[tex]F_{Y}[/tex] = F sin θ -----------------------(ii)
Now to the question;
Let the vector be A
Therefore;
The magnitude of vector A is A = 5.4m
The direction θ of A counterclockwise from the positive direction of the x-axis = 259°
(a) The x-component ([tex]A_{X}[/tex]) of the vector A is therefore given by;
[tex]A_{X}[/tex] = A cos θ ------------------------(iii)
Substitute the values of θ and A into equation (iii) as follows;
[tex]A_{X}[/tex] = 5.4 cos 259°
[tex]A_{X}[/tex] = 5.4 x (-0.1908)
[tex]A_{X}[/tex] = -1.030
Therefore, the x-component of the vector is -1.030m
(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;
[tex]A_{Y}[/tex] = A sin θ ------------------------(iv)
Substitute the values of θ and A into equation (iv) as follows;
[tex]A_{Y}[/tex] = 5.4 sin 259°
[tex]A_{Y}[/tex] = 5.4 x (-0.9816)
[tex]A_{Y}[/tex] = -5.301m
Therefore, the y-component of the vector is -5.301m
The dimension of a room is 10 feet high, 40 feet long and 30 feet wide. If ¼ point of pyridine is allowed to volatilize in this room at NTP with nor ventilation, what is the concentration in mg/m3 and ppm for pyridine in this room? Pyridine s.g. = 0.98
Answer:
Concentration in mg/m^3 = 3845.39mg/m^3
in PPM = 3.84539 ppm
Explanation:
From the dimensions, we compute the volume (V)
V = 10 x 40 x 30 = 12000ft^3
Volatilize volume = 1/4 *12000 = 3000ft^3 at NTP
Volume of pyridine in the room = ( 12000 - 3000) = 9000ft^3
converting to meters = 9000 * (0.3048)^3m^3
= 254. 85m^3
Computing concentration(C) = mass/s.g/volume
S.g converted to mg per meter cube = 0.98x1000= 0.98 * 10^3g/m^3 = 0.98 * 10^6mg/m^3
Hence concentration = 0.98 * 10^6/254.85 = 3845.39 mg/m^3
Parts per million denotes low concentration of a solution
1ppm = 1mg per liter , and 1 liter = 0.001m^3
Hence, this = 3845.39/1000 = 3.84539 PPM
A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.
Answer:
weight of the bullet is 0.0500 lb
velocity as it travels from block A to block B is 900 ft/s
Explanation:
given data
horizontal velocity = 1500 ft/s
mass block A = 6-lb
mass block B = 4.95 lb
blocks A velocity = 5 ft/s
blocks B velocity = 9 ft/s
solution
we apply here law of conservation of momentum that is
m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2 ................1
put here value and we get m
m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex] ..............2
m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]
m = 0.0500 lb
and
when here bullet is pass through the A block then moment is conserve that is
m × v(o) + m × v1 + m1 × v1 ............3
v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]
v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]
v1 = 900 ft/s
A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu ft. Establish the safe load if the number of blows to drive the last batch is 36, volume of base and plug is 27 cu ft, and the soil K value is 28.
Answer:
159.1 ton
Explanation:
The solution is shown in the attached file
Answer:
The safe load is 159 ton
Explanation:
The safe load is equal to:
[tex]L=\frac{WHBV^{2/3} }{K}[/tex]
Where:
W = weight of hammer = 2.5 ton
H = drop height = 22 ft
B = number of blows used to drive the last batch = 36/4 = 9 ft³
K = dimensionless constant = 28
V = uncompacted volume = 27 ft³
Replacing values:
[tex]L=\frac{2.5*22*9*(27^{2/3}) }{28} =159ton[/tex]
A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.30, and the coefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface
Answer:
The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N
Explanation:
To start sliding the box on the surface it must overcome its static frictional force under equilibrium condition
The net force on the box is
F - fs = 0
F = fs = us N = us mg
Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N
Horizontal force is defined as the forces that equal and opposite in the direction. The horizontal resultant force is always zero.
Given that:
Mass of box = 55 Kg
Coeeficient of Static friction = 0.30
Coefficient of kinetic friction = 0.20
The box if have to slide, then it would have to overcome the static force, such that:
F - Fs = 0
F = Fs
The force will be:
F = [tex]\rm \mu \times mass \times g[/tex]
F = [tex]0.3 \times 55 \times 9.8[/tex]
F = 161.7 Newton
Thus, the force required to slide the box is 161.7 Newton.
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Two identical wires A and B are subject to tension. The tension in wire A is 3 times larger than that in wire B. Find the ratio of the frequencies of the first harmonic in these two wires, fA1 / fB1.
Answer:
1.732
Explanation:
Let
Tension in wire B=T
Tension in wire A=3 T
We have to find the ratio of the frequencies of the first harmonic in these two wires.
When two wires are identical then the length of both wires are same.
Suppose, the length of each wire=l
Frequency=[tex]\frac{1}{2l}\sqrt{\frac{T}{\mu}}[/tex]
Where [tex]\mu=[/tex]Mass per unit length
Mass per unit length of both wires are same because the two wires are identical.
[tex]\mu_A=\mu_B[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_B}}}[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_A}}}=\sqrt{\frac{T_A}{T_B}}=\sqrt{\frac{3T}{T}}[/tex]
[tex]\frac{f_A}{f_B}=\sqrt 3=1.732[/tex]
The ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
The given parameters;
tension in wire A = Ttension in wire B = 3TThe frequency of first harmonic of each wire is calculated as follows;
[tex]F_ A = \frac{1}{2l} \sqrt{\frac{3T}{\mu} } \\\\F_B = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
l is the length of the wiresT is the tension on the wireμ is the mass per unit lengthThe ratio of the two frequencies is calculated as follows;
[tex]\frac{F_A}{F_B} = \frac{\frac{1}{2l} \sqrt{\frac{3T}{\mu} } }{\frac{1}{2l} \sqrt{\frac{T}{\mu} } } \\\\\frac{F_A}{F_B} = \sqrt{\frac{3T}{T} } \\\\\frac{F_A}{F_B} = \sqrt{3}[/tex]
Thus, the ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
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A cube has one corner at the origin and the opposite cornerat the point (L,L,L). The sides of the cube are parallel to thecoordinate planes. The electric field in and around the cube isgiven by E=(a+bx)x+cy.Part A Find the total electric flux phi E through the surface of the cube. Express your answer in terms of a,b,c and L. Phi E=Part B This part will be visible after youcomplete previous item(s). Part C What is the net charge q inside the cube? Express your answer in terms of a,b,c ,L,and epsilon. q=
Answer:
Explanation:
in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.
(flux=EAcos0 =EA)
flux through the face x face
Physics homework question answer, step 1, image 1
the flux through -x face
Physics homework question answer, step 1, image1
Step 2
the flux through y face,
Physics homework question answer, step 2, image 2
the flux through -y face,
Physics homework question answer, step 2, image 2
the flux through the z faces are zero,
therefore, the net flux is,
Physics homework question answer, step 2, image 3
...
Final answer:
The total electric flux IS ΦE = aL2 + (a + bL)L2 + cL3 net charge is q = ε0 ( aL2 + (a + bL)L2 + cL3 )
Explanation:
The question asks to find the total electric flux ΦE through the surface of a cube with an electric field given by E = (a + bx)x + cy, and to determine the net charge q inside the cube.
Part A: Total Electric Flux through the Cube's Surface
To determine the electric flux through the cube, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by the surface.
The electric flux ΦE through a surface is defined by the integral of the electric field E dotted with the differential area dA over the entire surface. Since the sides of the cube are parallel to the coordinate planes, the electric field components perpendicular to the x, y, and z faces will contribute to the flux.
In this case, only the x-component of the electric field (a + bx)x contributes to the flux through the surfaces perpendicular to the x-axis, which are at x = 0 and x = L.
The y-component cy of the electric field contributes to the flux through the surfaces perpendicular to the y-axis, which are at y = 0 and y = L. There is no z-component to the electric field, so the surfaces perpendicular to the z-axis will have no flux.
The total electric flux through the cube is thus:
ΦE = ∫( (a + bx) x + cy ) · dA
For the two faces perpendicular to the x-axis:
ΦE, x=0 = ∫( a x ) · dA = aL2 (at x = 0)
ΦE, x=L = ∫( (a + bL)x ) · dA = (a + bL)L2 (at x = L)
For the two faces perpendicular to the y-axis:
ΦE, y=0 = 0 (since cy is 0 at y = 0)
ΦE, y=L = ∫( cy ) · dA = cL3 (at y = L)
Summing these up:
ΦE = aL2 + (a + bL)L2 + cL3
Part C: Net Charge Inside the Cube
The net charge q inside the cube can be found using Gauss's Law, which states "the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (epsilon_0)."
q = ΦE ε0
Substituting the expression for ΦE we have:
q = ε0 ( aL2 + (a + bL)L2 + cL3 )