Answer:
The power generated by the windmill is approximately 1.364 MW
Explanation:
The diameter of the windmill, d = 30 m
The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s
The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s
The pressure at the inlet, P₁ = 2 atm
The pressure at the outlet, P₂ = 1 atm
The density of air, ρ = 1.2 kg/m³
The power obtained from the windmill, 'P', is given as follows;
[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]
Where;
[tex]g_c[/tex] = 1.0 kg/(N·s²)
A = Cross-sectional rea of the the windmill = π·D²/4 = π×(30 m)²/4 = 706.858347 m²
Plugging in the values, we get;
[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]
The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.
Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each other. What law of motion is this. (Newton's laws)
Answer:
forcing in act
Explanation:
Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C. (drag on jet / drag on transport)
Answer:
[tex]2.267[/tex]
Explanation:
Drag force is given by
[tex]F=\dfrac{1}{2}\rho Av^2C[/tex]
C = Drag coefficient is constant
A = Area is constant
[tex]v_1[/tex] = Velocity of the passenger jet = 1200 km/h = [tex]\dfrac{1200}{3.6}\ \text{m/s}[/tex]
[tex]v_2[/tex] = Velocity of the prop plane = [tex]\dfrac{1}{4}v_1[/tex]
[tex]\rho_1[/tex] = Density of the air where the jet was flying = [tex]0.38\ \text{kg/m}^3[/tex]
[tex]\rho_2[/tex] = Density of the air where the prop plane was flying = [tex]0.67\ \text{kg/m}^3[/tex]
[tex]F\propto \rho v^2[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267[/tex]
The ratio of the drag forces is [tex]2.267[/tex].
Jack weighs 170 lbs and is 72 inches tall. He is pulling horizontally on a door handle situated at his shoulder height. Actually, it is his body weight and lean that creates this pulling action (a hint). His center of mass while standing erect is 61 percent of his body height, measured from the floor upwards. The door handle is 60 inches above the ground, and again he is pulling purely horizontally on this handle.
If Jack's lean angle is 20 degrees and he is leaning back - pivoting about his heels, how much force does he apply to the door handle?
Include units in your answer, lbs.
Express your answer to the nearest 0.1 lbs.
Answer:
He is pulling horizontally on a door handle situated at his shoulder height. ... His Center Of Mass While Standing Erect Is 61 Percent Of His Body Height, Measured ... Actually, it is his body weight and lean that creates this pulling action (a hint).
72ibs
how does the strength of the forces that hold the basic particles of a substance together relate to the temperature at which the substance changes state
The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
an object of 5kg is attached to a rope of length 4m is Rotating horizontally at 8m/s horizontally 20m above the ground if the rope is suddenly cut what is the horizontal distance travelled by the object? Please guys help
Answer:
16 meters
Explanation:
When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.
First we need to find how much time de object take to reach at the ground.
VERTICALLY EQUATION:[tex]h(t)=h-v*t-\frac{g}{2} t^{2} \\[/tex]g=acceleration of gravity=10m/s²
v= vertical velocity =0m/s
h=vertical altitude =20m
We will find t such that h(t)=0
[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]
HORIZONTALLY EQUATION:*horizontally we do not have acceleration[tex]D(t)=v*t[/tex]v=horizontal velocity
D(t=2) is the horizontal distancetravelled by the object:
[tex]D(2)=8*2\\\\D(2)=16m[/tex]
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g
Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight
Answer:
D. Weight
Explanation:
Hope that helps:)
Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not equally sensitive to all frequencies of sound. In other words, a particular noise may sound louder or softer depending on the frequency of the sound wave being transmitted. Because of this variation, scientists have defined a unit of loudness, called a phon, to represent the intensity of sound waves with a frequency of 1000 Hz. A 60-phon sound is one that is perceived by the human ear to have the same loudness as a sound wave with an intensity of 60 dB and a frequency of 1000 Hz.
Required:
a. At approximately what frequency do most people perceive the least intense sounds? Answer numerically in hertz to two significant figures.
b. Normal conversation has a sound level of about 60 dB. How many times more intense must a 100-Hz sound be compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness?
Answer:
20 Hz
15.8 times
Explanation:
A
Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.
B
A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness
What are some technological limitations that currently prevent humans from traveling to distant planets?
Answer:
Propulsion system, antigravitational tech
Explanation:
Fuel is extremely inefficient and expensive not to mention it weighs a lot. You really only need to reach escape velocity to leave earth. The rest is just a little amount of boosting to alter course and slow down for landing. I couldn't really think of much. Once we have an antigravitational system then you could say the whole rocket is holding you back because the design would be different. Nobody really knows how to defy gravity but that would be a technolgical limitation for sure.
two small identical conducting spheres have charges of 2.0x10-9C and - 0.5x109 C respectively when they are placed 4cm apart, what is the force between them? If they are brought into contact and then separated by 4cm, what is the force between them?
Answer:
6
Explanation:
nothingnsbejejjdbsbzbawkje
PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 120oC while the other side contains 1 m3 of He gas at 500 kPa and 40oC. Assume the piston is made of 8 kg of copper initially at the average temperature of the two gases on both sides. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move
Answer:
The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".
Explanation:
Formula for calculating the mass in He:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]
Formula for calculating the mass in [tex]N_2[/tex]:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]
by using the temperature balancing the equation:
[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]
[tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]
A car travels 150 kilometers west in 3 hours. What is its average velocity?
Your answer:
150 km/hr
50 km/hr
50 km/hr west
150 km/hr west
Answer:
C= 50km/hr west
Explanation:
150/3= 50
Because it asks for velocity, make sure to include the direction as well.
How can you tell whether an object is neutral
or charged? What would you have to do to test
that object?
Answer:
The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral
Explanation:
something that orbiys other things in space
Answer: well we all orbit the sun all the planets do so the
SuN
Explanation: two words common sense
help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well
An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is
Answer:
You need to know the accuracy to which you can read the ruler:
Suppose that you can read the read the ruler to the nearest milimeter
A = L * W your calculated area of the rectangle
A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA
Or ΔA = L ΔW + W ΔL
Where we have subtracted A = L * W and the term ΔL * ΔA is very small
So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2
Then you report A = 10 cm^2 +- .7 cm^2 including the - sign for completeness
The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the
Equator bring cooler waters to some regions, while currents flowing away from the Equator bring warmer waters to other regions.
North
British
Isles
Askan
North Atlantic
Azor
U.S.A
California
Gulf Stream
Loop
n
Canbean
North Equatorial
North Equatorial CC
North Fuatorial
Equator
South Equatorial
Not
South Equatorial
Image courtesy of NOAA
Judging from the map, which region probably has cooler summers than it would without the effect of a nearby ocean current?
A the Central U.S.
B. the British Isles
C. the U.S. East Coast
D. the US West Coast
Answer:
d
Explanation:
the US West Coast region probably has cooler summers than it would without the effect of a nearby ocean current.
what is ocean current ?
ocean current can be defined as the horizontal movement of seawater which is produced by gravity, wind, and water density, it play an major role in the determination of climates of coastal regions.
The movement of ocean water is continuous which can be up three types such as Waves, Tides, Currents
The streams of water which flow continuously on the ocean surface in specific directions are called ocean currents, it affect the temperature of ocean water as Warm ocean currents increase the temperature whereas Cold ocean currents decrease the temperature.
The magnitude of the ocean currents is about few centimeters per second to as much as 4 metres per second and the intensity of the ocean currents generally decreases with increasing depth.
There are two types of ocean currents such as Warm Ocean Currents
and Cold Ocean Currents
For more details ocean current, visit
https://brainly.com/question/21654036
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A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
PLEASE HELP! WILL GIVE BRANILIEST TO FIRST REAL ANSWER If one marble is rolling three times as fast as a second marble of the same mass, the kinetic energy of the first marble is how many times larger when compared to the kinetic energy of the second marble?
a) 4
b) 9
c) 6
d) 3
(i already know its not 3)
Answer:
9
Explanation:
A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is?
Answer:
h over 2 dg
Explanation:
brainliest!!!!!!!
Select the correct answer.
The oceanic Nazca plate is being subducted beneath the continental South American plate. Which type of plate boundary is this?
OA continental-oceanic convergent
ОВ. oceanic-oceanic convergent
OC divergent
OD. strike-slip
ОЕ.
transform
Reset
Next
Answer:
A. continental-oceanic convergent
Explanation:
I knew it couldn't be B because it's oceanic and continental, not oceanic and oceanic.
Next, I noticed the word convergent, which implies "coming together" to me.
I looked it up and noticed the term convergent referred to a plate boundary where a plate slips under (subducted) another, so I knew it was A.
Hopefully, this helps you understand the question better. Have a great day!
Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
what is permittivity
Answer:
Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.
A 14.0-g wad of sticky clay is hurled horizontally at a 90-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact
Answer:the speed of the clay immediately before impact =72.58m/s
Explanation:
Given that
mass of the stick clay, M₁= 14.0 g = 0.014 kg
mass of the block ,M₂= 90 g = 0.09 kg
Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg
Also, distance, s = 7.50 m
coefficient of friction μ= 0.650
Acceleration due to gravity ,g = 9.8 m/s²
Using the Work- Energy theorem,
change in kinetic energy = work done
final kinetic energy(K₂) - initial kinetic energy(K₁) = force, F x coefficient of friction, μ x distance,s
The final kinetic energy is zero because after the impact, the block with the clay comes to a stop after 7.50m
kinetic energy =Work done
0.5 x m x v²=coefficient of friction, μ x force(F) x distance,s(Since force = m g )
0.5 x m x v²= μ x m x g x s
0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5
v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104
v²==95.55
V = 9.77 m/s
Using the conservation of momentum formulae where
M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V
Since V₂ which is the velocity of block is zero as the block is initially at rest, We now have that
M₁ V₁ = (M₁ + M₂ ) V
0.014 kg x V₁ = 0.104 x 9.77
V₁=0.104 x 9.77 / 0.014
V=72.58m/s
It's time to get a little more specific. Based on the velocity (Vx) graph for the car and the velocity data in the table, divide the total
motion of the car into rough time periods that tell a different "chapter" of the story for this car trip. In each of these time
periods, the car's velocity will be notably different from the previous period. Enter a brief description of the car's motion in each
period. The first one is done for you. Use it as an example to identify and describe the remaining time periods. Note: You can
define as many periods as you think appropriate.
s
B
1
U X
X х.
Font Sizes
А • А
E
E 를 들
E 3
Numbered list
Time Period
Motion Description
0.2 - 4.6 seconds increasing speed in positive direction
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
Scientists have investigated how quickly hoverflies start beating their wings when dropped both in complete darkness and in a lighted environment. Starting from rest, the insects were dropped from the top of a 50 - cm tall box. In the light, those flies that began flying 200 m s after being dropped avoided hitting the bottom of the box 87 % of the time, while those in the dark avoided hitting only 25 % of the time.
Required:
a. How far would a fly have fallen in the 200 ms before it began to beat its wings?
b. How long would it take for a fly to hit the bottom if it never began to fly? In seconds.
Answer:
Explanation:
a )
Hoverfly will fall with acceleration equal to g .
Initial velocity of fall of hoverflies u = 0
displacement ( vertical ) h = ?
time t = 0.2 s
acceleration due to gravity g = 9.8 m / s²
h = ut + 1/2 g t²
= 0 + .5 x 9.8 x .2²
= .196 m
= 19.6 cm
b )
Time taken to fall by 50 cm or 0.5 m under free fall from initial position .
.5 = 0 + .5 x 9.8 t²
t² = .1020
t = .319 s = 319 ms .
What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?
Answer:
speed = 10.1215 m/s
Explanation:
speed = distance / time
speed = 100 / 9.88 = 10.1215 m/s
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposing each other) in a circuit. If you wanted to add in a capacitor to charge it from the batteries, would you be able to get more charge onto the capacitor or less charge, than if there was only one battery. (hint: start this problem by aligning the batteries positive to negative, and think of it from conservation of energy perspective).
Answer:
Answer is explained in the explanation section below.
Explanation:
This question is very basic and easy. The answer to this question is:
Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.
Reasoning:
If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.
As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.
In picture 1, heat is flowing from the ____ to the _____ In picture 2, heat is flowing from the _______ to the ____
Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid
Explanation:
I don't know if I answered correctly, if not I can provide another answer
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
