Answer:
The correct choice is B & E.
Explanation:
A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.
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A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
0.99Hz
Explanation:
Using F= -mx ( spring force)
At equilibrium the gravitational force will be balanced by the spring force so mg= kx
K= mg/ 0.25 N/m
But
Frequency f= 1/2pi √g/0.25
Frequency is 0.99Hz
The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz
Frequency of oscillation based problem:What information do we have?
Length starched = 2.5 cm
F = Kx
We know that
F = mg
So,
mg = Kx
K/m = g/x
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]
Frequency of oscillation = 3.15 Hz
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Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.
Answer:
A) 1.5 V
B) 4.5 V
Explanation:
A) Batteries in parallel have the same voltage as an individual battery.
V = 1.5 V
B) Batteries in series have a voltage equal to the sum of the individual batteries.
V = 1.5 V + 1.5 V + 1.5 V
V = 4.5 V
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
Ratio of the speed of light in a vacuum to the speed of light in a medium Rule for how light is refracted at the boundary between two materials Process that occurs when the angle of incidence is greater than the critical angle
Answer:
TOTAL INTERNAL REFLECTION
Explanation:
Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.
For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.
Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION
A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?
Answer:
50 cm long
When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.4 GHz, and as they do so they emit 2.4 GHz electromagnetic waves. What is the strength of the magnetic field?
Answer:
The magnetic field is 0.0857 T.
Explanation:
The electrons orbit the magnetic field with a centripetal force equal to
F = [tex]\frac{mv^{2} }{r}[/tex]
also, the force on an electron in a magnetic field is gotten as
F = Bqv
equating this two equations give
[tex]\frac{mv^{2} }{r}[/tex] = Bqv
mv/r = Bq
where m is the mass of the electron = 9.11 x 10^-31 kg
v is the the linear speed of the electron
B is the magnetic field on the electron
r is the radius of the orbital movement
q is the charge on an electron = 1.602 x 10^-19 C
but, the linear speed v = ωr
where ω is the angular speed of the electron
substituting into equation above, we have
mωr/r = Bq
which reduces to
mω = Bq
finally, w know that the angular speed is related to the frequency of the electron by
ω = 2πf
we then finally have
2mπf = Bq
where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz
substituting values into the equation, we have
2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19
B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T
= 85.7 mT
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?
Answer:
c. 70 Ω
Explanation:
The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:
Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)
Req = R/2 + 2R/2 + 4R/2
Req = 3.5R
Req = 70Ω
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.
a. True
b. False
Answer:
True or False
Explanation:
Because.....
easy 50% chance you are right
how does a system naturally change over time
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?
Answer:
7.86×10⁻⁷ m
Explanation:
Using,
v = λf.................. Equation 1
Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.
make λ the subject of the equation
λ = v/f............... Equation 2
Note: All electromagnetic wave have the same speed which is 3×10⁸ m/s.
Given: f = 3.818×10¹⁴ Hz
Constant: v = 3×10⁸ m/s
Substitute these values into equation 2
λ = 3×10⁸/3.818×10¹⁴
λ = 7.86×10⁻⁷ m
Hence the wavelength of the electromagnetic radiation is 7.86×10⁻⁷ m
The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]
Given the following data:
Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]Scientific data:
Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]
To determine the wavelength of this electromagnetic radiation:
Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;
[tex]Wavelength = \frac{Speed }{frequency}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]
Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]
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A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?
Explanation:
Using Equations of Motion :
[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]
Height = 0 * 4 + 4.9 * 16
Height = 78.4 m
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form _______. A. pure elements B. molecules C. metals D. the periodic table
Answer:
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form pure elements
The answer is option A
Answer:
its molecues
Explanation:
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?
Answer:
The time taken is [tex]\Delta t = 1.5 \ s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 1.25 \ kg[/tex]
The time taken to make the first complete revolution is t= 3.60 s
The displacement of the first complete revolution is [tex]\theta = 1 rev = 2 \pi \ radian[/tex]
Generally the displacement for one complete revolution is mathematically represented as
[tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]
Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]
[tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]
[tex]\alpha = 0.9698 \ s[/tex]
Now the displacement for two complete revolution is
[tex]\theta_2 = 2 * 2\pi[/tex]
[tex]\theta_2 = 4\pi[/tex]
Generally the displacement for two complete revolution is mathematically represented as
[tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]
=> [tex]t^2 = 25.9187[/tex]
=> [tex]t= 5.1 \ s[/tex]
So
The time taken to complete the next oscillation is mathematically evaluated as
[tex]\Delta t = t_2 - t[/tex]
substituting values
[tex]\Delta t = 5.1 - 3.60[/tex]
[tex]\Delta t = 1.5 \ s[/tex]
The time for the ball to complete the next revolution is 1.5 s.
The given parameters;
mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2πThe constant angular acceleration of the ball is calculated as follows;
[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]
The time to complete the next revolution is calculated as follows;
[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]
[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]
Thus, the time for the ball to complete the next revolution is 1.5 s.
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An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)
Answer:
10.4⁰ and 169.6⁰Explanation:
The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;
q is the charge on the electron
v is the velocity of the electron
B is the magnetic field strength
θ is the angle that the velocity of the electron make with the magnetic field.
Given parameters
F = 1.40*10⁻¹⁶ N
q = 1.6*10⁻¹⁹C
v = 3.94*10³m/s
B = 1.23T
Required
Angle that the velocity of the electron make with the magnetic field
Substituting the given parameters into the formula:
1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ
sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶
sinθ = 1.40/7.75392
sinθ = 0.1806
θ = sin⁻¹0.1806
θ₁ = 10.4⁰
Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁
θ₂ = 180-10.4
θ₂ = 169.6⁰
Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circumstances. (Note that currents above 10 mA lead to involuntarily muscle contraction.)
(a) if she is standing on a rubber mat and offers a total resistance of 300kΩ
(b) if she is standing barefoot on wet grass and has a resistance of only 4000kΩ
Answer:
A) 0.4 mA
B) 0.03 mA
Explanation:
Given that
voltage source, V = 120 V
to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.
mathematically, Ohms Law, V = IR
V = Voltage
I = Current
R = Resistance
from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have
120 = I * 300*10^3 Ω
making I the subject of the formula,
I = 120 / 300000
I = 0.0004 A
I = 0.4 mA
Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.
B, we have Resistance, R = 4000kΩ
Substituting like in part A, we have
120 = I * 4000*10^3 Ω
I = 120 / 4000000
I = 0.00003 A
I = 0.03 mA
This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA
The current through a person will be:
a) 0.4 mA
b) 0.03 mA
Given:
Voltage, V = 120 V
Ohm's Law:It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.
Ohms Law, V = I*R
where,
V = Voltage
I = Current
R = Resistance
a)
Given: Resistance= 300kΩ
[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]
Thus, current will be, I = 0.4 mA
b)
Given: R = 4000kΩ
[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]
Thus, current will be, I = 0.03 mA
From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.
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UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".
Kinetic energyAccording to the question,
Speed of light, c = 3 × 10⁸ m/s
Wavelength, λ = 120 nm or,
= 1.2 × 10⁻⁷ m
Plank's Constant, h = 6.626 × 10⁻³⁴ J.s
Now,
The energy of photon will be:
→ E = [tex]\frac{hc}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]
= [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]
= 10.35 eV
By using Einstein's Photoelectric equation,
Energy of Photon = Work function + K.E
10.35 = 4.82 + K.E
K.E = 10.35 - 4.82
= 5.53 eV or,
= 8.85 × 10⁻¹⁹ J
Thus the response above is correct.
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Please help!
Much appreciated!
Answer:
your question answer is 22°
The two metallic strips that constitute some thermostats must differ in:_______
A. length
B. thickness
C. mass
D. rate at which they conduct heat
E. coefficient of linear expansion
Answer:
E. Coefficient of linear expansion
A homeowner purchases insulation for her attic rated at R-15. She wants the attic insulated to R-30. If the insulation she purchased is 10 cm thick, what thickness does she need to use
Answer:
she need to use 20 cm thick
Explanation:
given data
wants the attic insulated = R-30
purchased = 10 cm thick
solution
as per given we can say that
10 cm is for the R 15
but she want for R 30
so
R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]
R 30 thickness = 20 cm
so she need to use 20 cm thick
There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,
1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N
Explanation:
Pressure is the same for both plungers.
P = P
F / A = F / A
F / (¼ π d²) = F / (¼ π d²)
F / d² = F / d²
5 N / (0.05 m)² = F / (1 m)²
F = 2000 N
None of the options are correct.
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q
Answer:
Upper plate Q/3
Lower plate 2Q/3
Explanation:
See attached file
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
