Answer:
191 cm³
Explanation:
We'll begin by converting celsius temperature to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 20 °C
Initial temperature (T₁) = 20 °C + 273
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 100 °C
Final temperature (T₂) = 100 °C + 273
Final temperature (T₂) = 373 K
Finally, we shall determine the final volume of the air. This can be obtained as follow:
Initial volume (V₁) = 150 cm³
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 373 K
Pressure = constant
Final volume (V₂) =?
V₁/T₁ = V₂/T₂
150 / 293 = V₂ / 373
Cross multiply
293 × V₂ = 150 × 373
293 × V₂ = 55950
Divide both side by 293
V₂ = 55950 / 293
V₂ ≈ 191 cm³
Thus, the final volume of the air is 191 cm³
Why is it that, when we observe an extragalactic source whose diameter is about one lightday, we are unlikely to see fluctuations in light output in times shorter than about one day
The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.
How to observe extragalactic sources whose diameter is about one light day?When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.
For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.
In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.
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como calcular la velocidad un atleta en los 100 metros planos?
Answer:
Explanation:
9ooooo
6. A transverse periodic wave on a string with a linear density of 0.200 kg/m is described by the following equation: y = 0.08 sin(469t – 28.0x), where x and y are in meters and t is in seconds. What is the tension in the string? A) 3.99 N B) 32.5 N C) 56.1 N D) 65.8 N
Answer:
T = 56.11 N
Explanation:
Given that,
The equation of a wave is :
y = 0.08 sin(469t – 28.0x),
where x and y are in meters and t is in seconds
The linear mass density of the wave = 0.2 kg/m
The speed of wave is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Also,
[tex]v=\dfrac{\omega}{k}[/tex]
We have,
[tex]k=469\ and\ \omega=28[/tex]
Put all the values,
[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]
Put all the values,
[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]
So, the tension in the string is 56.11 N.
Two horizontal forces are acting on a box. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that = +5.9 N and the mass of the box is 3.6 kg. Find the magnitude and direction of when the acceleration of the box is +7.1 m/s^2.
Answer:
sorry I don't know I am only in 7th grade
A television of mass 8 kg sits on a table. The coefficient of static friction
between the table and the television is 0.48. What is the minimum applied
force that will cause the television to slide?
O A. 38 N
O B. 62 N
O C. 78 N
D. 55 N
The television has weight (8 kg) g = 78.4 N, and the magnitude of the normal force between the table and television would be the same, 78.4 N. This mean the maximum magnitude of static friction between the table and television is
0.48 (78.4 N) ≈ 37.6 N ≈ 38 N
and this is the minimum required force needed to get the television to slide.
how do you use the coefficient to calculate the number of atoms in each molecule?
Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
Which of the statements below are TRUE! Group of answer choices The carbon rod in batteries react to form a carbon cation. A good car battery gives you a little bit of power for a long period of time. A good car battery gives you a lot of power in a short period of time. The carbon rod in batteries is used as an inert electrode.
Answer:
The carbon rod in batteries is used as an inert electrode
Explanation:
A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.
It contains a positive terminal and a negative terminal.
The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.
Thus the true statement is :
The carbon rod in batteries is used as an inert electrode.
Testing shows that a sample of wood from an artifact contains 50% of the original amount of carbon-14. Given that the half-life of carbon-14 is 5730
years, how old is the artifact?
A. 22,920 years
B. 11,460 years
C. 5730 years
D. 2865 years
Answer:
C. 5730 years
Explanation:
N(t) = N(0)e^-kt
The half-life is T = 5730 years,
e^-kT = 1/2
→ k = - ln(1/2) / T
→ - ln(1/2) / 5730
→ 1.209681 x 10^-4 years^-1
The amount present dropped to 50%.
Then one half-life has elapsed, so the age is 5730 years.
i didnt want my question public i made a mistake i want it taken down
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
A) 8.03 x 10^16 nuclei
B) 4.01 x 10^16 nuclei
C) 2.02 x 10^16 nuclei
D) 1.61 x 10^17 nuclei
OPTION C is the correct answer.
The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.
What is half-life?The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.
The isotope decay of an atom is given by the equation:
ln [A] = -kt + ln [A]₀
The rate constant, k is:
k = ln 2 / Half life
k = ln 2 / 4.96 x 10³
k = 1.40 × 10⁻⁴ s⁻¹
t = 1.98 x 10⁴
[A]₀ = 3.21 x 10¹⁷
ln [A] = -1.40 × 10⁻⁴ × 1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538
[A] = 2.02 x 10¹⁶ nuclei
Thus the correct option is C.
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A ball is thrown with an initial velocity of 30.0 m/s and makes an angle of
30.0° with the ground. Find the
A.Horizontal Distance
B.Maximum Height
C.Total Time The Ball is Traveling
Statements imply it is thrown with velocity 30cos30° horizontally and 30sin30° vertically.
Vertically:
Total time taken = 2 x time to go up
= 2(v - u)/a
= 0 - 30sin30°)/(-g)
= 30/g
Therefore, it would travel 30/g sec in horizontal direction as well.
Horizontally :
Distance = horizontal speed x time
= 30cos30° (30/g)
= 450√3 /g
If g = 10, distance is 45√3 m.
Vertically,
Distance = vert. speed x (time of flight/2)
= 30sin30° x (30/g)/2
= 90 m.
Time taken = 30/g = 3 sec
(c) Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0= 1.68×10-21 J and R0= 3.82×10-10 m. Find the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
Explanation:
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
The formula for calculating the elastic potential energy is:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
By rearrangement and using (K) as the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the formula used to calculate the frequency of the small oscillation is:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom
assuming
m = 1.66 × 10⁻²⁶ kg
Then:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
How long will be required for a car to go from a speed of 27.0m/s to a speed of 32.0m/s if the acceleration is 2 3.0m / s ?
Answer:
0.217seconds is the answer
The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 18 cm above the waterline and a horizontal distance of 28 cm away. The fish aims its stream at an angle of 39° from the waterline.
Required:
Determine the height above the waterline that the stream reaches at the horizontal position of the insect.
Answer:
The fish gobbles the mosquito at height 18 cm.
Explanation:
Initial velocity, u = 3.7 m/s
horizontal distance, d = 28 cm
Angle, A = 39 degree
Let the time is t.
Horizontal distance = horizontal velocity x time
d = u cos A x t
0.28 = 3.7 cos 39 x t
t = 0.097 s
Let the height is h.
Use the second equation of motion
[tex]h =u t-0.5 gt^2\\\\h= u sin A t - 0.5 gt^2\\\\h= 3.7 sin 39 \times 0.097 - 0.5\times 9.8\times 0.097\times0.097\\\\h =0.226 -0.046 \\\\h=0.18 m=18 cm[/tex]
scripture union was founded by who in what year
Answer:
Josiah Spiers in 1867 was when scripture union was founded
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
At a particular instant, a proton, far from all other objects, is located at the origin. The proton is traveling with velocity (-3 x 106,0,0)m/s. Consider the electric and magnetic fields at observation point (9 x 10-10,2 x 10-10,0)m caused by this proton.
What is the electric field at the observation point?
What is the magnetic field at the observation point?
Answer:
The electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
The magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
Explanation:
Calculating the electric field at the observation point:We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m[/tex]
The equation used to calculate the electric field follows:
[tex]\vec{E}=\frac{kq\hat{r}}{r^2}[/tex]
OR
[tex]\vec{E}=\frac{kq}{r^2}\frac{\vec{r}}{|\vec{r}|}[/tex]
We know:
[tex]|\vec{r}|=r[/tex]
[tex]q=1.6\times 10^{-19}C[/tex]
So, the equation becomes:
[tex]\vec{E}=\frac{kq(\vec{r})}{r^3}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\vec{E}=\frac{(9\times 10^9}(1.6\times 10^{-18})(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})}{\left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{E}=(1.84\times 10^{18}(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})\\\\\vec{E}=1.65\times 10^{9}\hat{i}+3.68\times 10^{8}\hat{j}[/tex]
Hence, the electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
Calculating the magnetic field at the observation point:The magnetic field due to moving charge is given by:
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \hat{r}}{r^2}[/tex]
OR
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \vec{r}}{r^3}[/tex] .....(2)
We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m\\\\\vec{v}=-3\times 10^6\hat{i}[/tex]
Putting values in equation 2, we get:
[tex]\vec{B}=\frac{\mu_o\times (1.6\times 10^{-19})\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]}{4\pi\times \left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{B}=(20.42)\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]\\\\\vec{B}=-1.23\times 10^{-14}\hat{k}[/tex]
Hence, the magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
Answer:
For (a): The chemical shift is [tex]2.08\delta[/tex]
For (b): The chemical shift is [tex]9.85\delta[/tex]
For (c): The chemical shift is [tex]7.5\delta[/tex]
Explanation:
To calculate the chemical shift, we use the equation:
[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]
Given value of spectrometer frequency = 200 MHz
For (a):Given peak position = 416 Hz
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]
For (b):Given peak position = [tex]1.97\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]
For (c):Given peak position = [tex]1.50\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]
Suppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm
Answer:
0.035 J
Explanation:
Applying,
W = ke²/2.............. Equation 1
Where W = workdone by the stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = 2W/e²............... Equation 2
From the question
Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m
Substitute these values into equation 2
k = (2×2)/(0.15²)
k = 177.78 N/m
Hence, work need to stretch the spring from 33 cm to 35 cm
therefore,
e = 35-33 = 2 cm = 0.02 m
Substitute into equation 1
W = 177.78(0.02²)/2
W = 0.035 J
A vertical spring with a spring constant of 2.00 N/m has a 0.30-kg mass attached to it, and the mass moves in a medium with a damping constant of 0.025 kg/s. The mass is released from rest at a position 5.0 cm from the equilibrium position. How long will it take for the amplitude to decrease to 2.5 m?
Answer:
17 seconds
Explanation:
Given that:
The mass attached to the spring (m) = 0.30 kg
The spring constant (k) = 2.00 N/m
The damping constant (b) = 0.025 kg/s
The initial distance [tex]x_o[/tex] = 5.0 cm
The initial final amplitude [tex]A_f[/tex] = 2.5 cm and not 2.5 m, please note the mistake, if it is 2.5 m, our time taken will be -93.7 sec, and we do not want a negative time value.
To start with the angular frequency damping using the formula:
[tex]\omega_{\gamma}= \dfrac{b}{2m}[/tex]
[tex]\omega_{\gamma}= \dfrac{0.025 \ kg/s}{2(0.3 \ kg)}[/tex]
[tex]\omega_{\gamma}=4.167 \times 10^{-2} \ s^{-1}[/tex]
In the absence of damping, the angular frequency is:
[tex]\omega_o = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega_o = \sqrt{\dfrac{2 \ N/m}{0.3 kg}} \\\\\omega_o = 2.581 \ s^{-1}[/tex]
The initial amplitude oscillation can be computed by using the formula:
[tex]A_i = e^{-\omega_{\gamma}t} x_o \sqrt{\dfrac{\omega_o^2}{\omega_o^2-\omega_f^2}}[/tex]
[tex]A_i = e^{-\omega_{\gamma}0} (5.0 \ cm) \sqrt{\dfrac{2.581^2}{2.581^2-(4.167*10^{-2})^2}}[/tex]
[tex]A_i = 5.0006 \ cm \\ \\ A_i = 5.001 \ cm[/tex]
The final amplitude, as well as the initial amplitude, can be illustrated by using the relation:
[tex]A_f = e^{-\omega_{\gamma}t}A_i\\ \\ e^{-\omega_{\gamma}t} = \dfrac{2. 5 \ cm}{5.001 cm}\\ \\ = 0.4999\\ \\ \implies -\omega_{\gamma}t_f = \mathsf{In (0.4999)} \\ \\ t_f = \dfrac{\mathsf{-In (0.4999)}}{4.167*10^{-2} \ s^{-1}} \\ \\[/tex]
[tex]t_f = 16.64 \ sec \\ \\ \mathbf{t_f \simeq 17 sec}[/tex]
Question 6 of 10
A 2 kg vase is sitting on a 1 m high table. What is the vase's potential energy?
A. 196 J
B. 2J
C. 0.5 J
D. 19.6 J
SUBMIT
Answer: Choice D) 19.6 J
=======================================
Work Shown:
m = 2 kg = mass
g = 9.8 m/s^2 = acceleration of gravity (approximate)
h = 1 m = height
---------
PE = potential energy
PE = m*g*h
PE = 2*9.8*1
PE = 19.6 J
The vase has approximately 19.6 Joules of potential energy.
We didn't have to make any conversions because the unit Joule is equivalent to the more complicated unit of kg*m^2/s^2 so it only involves kilograms, meters and seconds. If however the mass was given in say grams (instead of kg), then you'd need to convert to kg.
Directions: Analyze and illustrate the given problems. Show your mathematical equations.
1. How much work is done when you lift an object that weighs 180 N to a height of 12 meters?
2. A cylindrical container having a mass of 50 kg is being pushed up an inclined plane. How much work
is done on the container when it is 6 meters above the floor?
3. How much work do you do to a 16-N rock that you carry horizontally across a 4m room?
Answer:
1. 2160 J
2. 2940 J
3. 64 J
Explanation:
1. Determination of the work done.
Weight (W) = 180 N
Height (h) = 12 m
Workdone =?
Wd = W × h
Wd = 180 × 12
Wd = 2160 J
Thus, the Workdone is 2160 J
2. Determination of the work done.
Mass (m) = 50 Kg
Height (h) = 6 m
Acceleration due to gravity (g) = 9.8 m/s²
Workdone =?
Wd = mgh
Wd = 50 × 9.8 × 6
Wd = 2940 J
Thus, the Workdone is 2940 J
3. Determination of the work done.
Force (F) = 16 N
Distance (d) = 4 m
Workdone =?
Wd = F × d
Wd = 16 × 4
Wd = 64 J
Thus, the Workdone is 64 J
Kilometer is a unit of length where as kilogram is a unit of mass
By George, you've nailed it, Stacy !
That's a fact, uh huh.
Truer words were never written.
Your statement is one of unquestionable veracity.
The pure truthiness of it cannot be denied.
Was there a question you wanted to ask ?
Can someone help me
Answer:
Explanation:before the phase change the substance is a particle.
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting
Answer:
the person is sitting 1.5 m from the left end of the board
Explanation:
Given the data in the question;
Wb = 125 N
Wm = 500 N
T₂ = 250 N
Now, we know that;
T₁ + T₂ = Wb + Wm
T₁ + 250 = 125 + 500
T₁ = 125 + 500 - 250
T₁ = 375 N
so tension of the left chain is 375 N.
Now, taking torque about the left end
500 × d + 125 × 2 = 250 × 4
500d + 250 = 1000
500d = 1000 - 250
500d = 750
d = 750 / 500
d = 1.5 m
Therefore, the person is sitting 1.5 m from the left end of the board.
Write the relation connecting Celsius scale and Fahrenheit scale of temperature
Answer: See explanation
Explanation:
Celsius and Fahrenheit are the scales that are used in the measurement of temperature. Celcius is also refered to as centigrade. The relation that exist between Celsius and Fahrenheit is typically proportional.
The conversion from Celsius to Fahrenheit is expressed as:
F = (9/5 × C) + 32.
The conversion from Fahrenheit to Celsius is expressed as:
C = 5/9(F - 32)
For example to convert 100°C to Fahrenheit will be:
F = (9/5 × C) + 32.
F = (9/5 × 100) + 32
F = 180 + 32
F = 212°F
2.
A Velocidade Escalar de um automóvel aumenta de 36km/h para 108km/h em 10 segundos.
Determina a sua aceleração media.
Answer:
Aceleración, a = 2 m/s²
Explanation:
Dados los siguientes datos;
Velocidad inicial = 108 km/h
Tiempo = 10 segundos
Velocidad final = 36 km/h
To find the average acceleration;
Conversión:
36 km/h to meters per seconds = 36*1000/3600 = 10 m/s
108 km/h to meters per seconds = 108*1000/3600 = 30 m/s
I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;
[tex] V = U + at[/tex]
Dónde;
V es la velocidad final.
U es la velocidad inicial.
a es la aceleración.
t es el tiempo medido en segundos.
Sustituyendo en la fórmula, tenemos;
[tex] 30 = 10 + a*10 [/tex]
[tex] 30 = 10 + 10a [/tex]
[tex] 10a = 30 - 10 [/tex]
[tex] 10a = 20 [/tex]
[tex] Aceleracion = \frac{20}{10}[/tex]
Aceleración, a = 2 m/s²
If you blow air between a pair of closely-spaced Ping-Pong balls suspended by strings, the balls will swing
A) toward each other.
B) apart from each other.
C) away from the air stream.
Answer:
c
Explanation:
A uniformly charged, straight filament 4.95 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder 1.65 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find:
a. the electric field at the surface of the cylinder
b. the total electric flux through the cylinder
Answer:
The electric field at the cylinder surface = 80.19 kN/C
Electric flux via the cylinder [tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]
Explanation:
Given that:
For the filament
The length = 4.95 m
The charge = 2.00 µC
The charge per unit length for the filament can be computed as:
[tex]\lambda = \dfrac{q}{l}[/tex]
[tex]\lambda = \dfrac{2}{4.5}\mu C/m[/tex]
Using Gauss's law:
[tex]\phi_E = \oint E^{\to}*dA^{\to}[/tex]---- (1)
where;
electric flux = [tex]\phi_E[/tex]
permittivity of free space = [tex]\varepsilon_o[/tex]
electric field = E
surface area = dA
However, the electric flux [tex]\phi_E[/tex] via the cylinder can be expressed as:
[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]
Equation (1) can now be rewritten as:
[tex]\dfrac{\lambda l'}{\varepsilon_o} = \oint E^{\to}*d(2 \pi rl')[/tex]
[tex]|E| =(\dfrac{\lambda l'}{\varepsilon_o 2 \pi rl' })[/tex]
replacing the values into the above equation:
[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \mu C/m) (1.65 \ cm)}{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (10 \ cm )(1.65 cm) })[/tex]
[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \times 10^{-6} C/m) }{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (0.1 \ m ) })[/tex]
[tex]\mathbf{|E| =80.19 \ kN/C}[/tex]
Thus, the electric field at the cylinder surface = 80.19 kN/C
The electric flux now is calculated using the said formula:
[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]
[tex]\phi_E = \dfrac{(\dfrac{2}{4.5}\times 10^{-6} \ C)(0.0165 \ m)}{8.85 \times 10^{-12} \ C^2/Nm^2}[/tex]
[tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]
what recommendations and coclusions can yiu make on the issue of human rights violation to Department of education ?
I recommend that they chill out. After that, they can do a web search on the phrase "human rights." They will learn that it describes each particular person's political objectives, at least those who claim to be morally superior to everyone else.