Answers
Answer:
ACD
Explanation:
according to it's description above
The wave is a mechanical wave and can also be sound wave also the wave is a longitudinal wave. The correct option is A, C, and D.
What is mechanical wave?A mechanical wave is one whose energy cannot be transmitted via a vacuum. To transfer their energy from one place to another, mechanical waves need a medium.
A mechanical wave is something like a sound wave. A vacuum can not be traversed by sound waves.
Waves like mechanical waves require a medium to travel through. Non-mechanical waves are those that can travel through any medium.
Mechanical waves include, but are not limited to, sound waves, water waves, and seismic waves.
The three claims are accurate as follows: The wave is a longitudinal wave and can also be a mechanical wave or a sound wave.
Thus, the correct option is A, C, and D.
For more details regarding mechanical waves, visit:
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Your question seems incomplete, the missing image is:
Related Questions
please answer correctly
Answers
Answer:
616000 J.
Explanation:
Bi. Determination of the work done.
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:
Workdone = Force × distance
Wd = F × s
With the above formula, we can obtain the Workdone as follow:
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Wd = F × s
Wd = 220 × 2800
Wd = 616000 J.
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. Cengel, Yunus; Cengel, Yunus. Thermodynamics: An Engineering Approach (p. 98). McGraw-Hill Higher Education. Kindle Edition.
Answers
Answer:
1767Kw
Explanation:
Velocity of wind = 10 m/s
diameter of the blades= 60m
ρ= air density = 1.25 kg/m3
Acceleration due to gravity= 9.81 m/s^2
Mechanical energy of the wind can be calculated using the expression below
Energy= (e*m)
= ρ V A e............eqn(1)
Where A= area
ρ= air density
e= wind energy per unit mass of air
e= (v^2)/2..........eqn(2)
If we substitute the values into eqn (2) we have
e= [(10)^2]/2
=50J/Kg
But Area=A= (πd^2)/4
Area= ( π× 60^2)/4
Area=2827.8m^2
If we input substitute the values into eqn (1) we have
Energy= 1.25 ×10 × 50×2827.8
=1767145.7W
We can convert to kilo watt
=1767145.7W/ 1000
= 1767Kw
Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answers
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL. The distance between the two locations is 2000 km and the transmission lines are made of 10 cm diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) 12 V I0 _
Answers
Answer:
The answer is below
Explanation:
The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m
The resistance (R) of transmission line is given as:
Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4
[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]
Power = [tex]\frac{V_L^2}{R_L}[/tex]
Power = 10 MW = 10 * 10⁶ W
[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]
[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]
a) Since there are two tranmission lines, the power consumed by the lines is:
[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]
b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W
Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%
what is a vector quantity?
Answers
Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.
The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.
Which number label represents the cell membrane?
1
2
4
6
(this is middle school science)
Answers
Answer:
1. cell membrane
2. golgi body
3. mitochondrion
4. cytoplasm
5. nucleolus
6. nucleus
Explanation:
The correct answer to this question is Option A; 6.
Why?
In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.
~Thank you~
Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].
One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.
a. Stretch
b. Vibrate
c. Contract
d. Expand
Answers
Answer:
The correct answer is option D.
Explanation:
It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.
I hope this helps.
A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 volts, how many volts will the secondary coil have?
242 volts
288
353
467
Answers
Answer:
467 volts
Explanation:
Vs/Vp = Ns/Np
Vs = Ns/Np × Vp
Vs = 350/150 × 200 = 7/3 × 200
Vs = 467 volts
ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law say about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.
Answers
Answer:
A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.
(Hope this helps! Btw, I am the first to answer.)
The string will break if the tension in
it exceeds 0.180 N. What is the
smallest possible value of d (in cm)
before the string breaks?
Answers
Answer:
define d first?
you need to list more variables
Answer:
list more valuable unit
what is the mathematical formula associated with newton's 2nd law of motion?
Answers
Answer:
F= m x a
Explanation:
Force (f) = mass (m) x acceleration (a)
How do you classify flammable liquid to gas and solid?
Answers
Answer:
A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.Add to that a gaggle of state, regional, and local authorities; and we feel compelled to begin this blog entry with our favorite caveat: get expert advice before deciding what to do with that rusting drum of stale gasoline out back.Explanation:
I hope it help you;)
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride
Answers
Answer:
Answer is explained in the explanation section below.
Explanation:
In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.
a) You pass through the highest point:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.
And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.
b) You pass through the lowest point of the ride:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.
And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.
What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?
Answers
Answer:
86.33m/s^2
Explanation:
Acceleration = Force/Mass
= 25.9/0.3
= 86.33
please help me I'm begging you Define and give examples of elements and compounds the structure of atoms (electrons, neutrons, and protons)
Answers
Water - Formula: H2O = Hydrogen2 + Oxygen. ...
Hydrogen Peroxide - Formula: H2O2 = Hydrogen2 + Oxygen2 ...
Salt - Formula: NaCl = Sodium + Chlorine. ...
Baking Soda - Formula: NaHCO3 = Sodium + Hydrogen + Carbon + Oxygen3 ...
Octane - Formula: C8H18 = Carbon8 + Hydrogen18 Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged). The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure.
At what tempreture will the of and oC
be the same
Answers
Answer:
-40 degreesTo find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
Explanation:
Hope it is helpful.....
Answer:
To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answers
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
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
Answers
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
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
Answers
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]
For a given substance, the molecules
move fastest when the substance is
Answers
Answer:GAS
Explanation:
Fill in the question
Answers
5) 30 seconds
6) 1.83m/s
a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?
Answers
Answer:
2.5 m/s^2
Explanation:
First, convert 90 km/hr into m/s:
90/3.6 = 25 m/s
vf = final velocity = 25 m/s
vi = initial velocity = 0 m/s
t = time = 10 seconds
a = acceleration, unknown
Then, find a using the following equation:
(vf - vi)/t = a
(25 m/s)/10 s = 2.5 m/s^2
a = 2.5 m/s^2
Hope this helps!! :)
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
Answers
If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow
Answers
Answer:fastest,same,slow down,opposite,slow
Explanation:
A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.
Velocity of a satellite around the planet.If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).
While moving towards the Earth (along the path from D to A) there is a component of force in the same direction as the motion; this causes the satellite to speed up.
While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.
Learn more about motion of satellite here: https://brainly.com/question/25721729
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Which example describes a nonrenewable resource?
Answers
Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.
A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?
Answers
AnswerHM???
Explanation:
I dONT KNOW
A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV
Answers
Answer:
The magnitude of the potential difference is 900 V.
Explanation:
Given;
mass of the bead, m = 4.0 g = 0.004 kg
charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C
final velocity of the bead, v = 3 m/s
What is the magnitude of the potential difference V?
Apply the principle of conservation of energy;
The electric potential energy at the beginning is equal to kinetic energy of the bead at the end of the journey.
qV = ¹/₂mv²
[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]
Therefore, the magnitude of the potential difference is 900 V.
The magnitude of the potential difference (V) is equal to 900 Volts.
Given the following data:
Mass of bead = 4.0 g to kg = 0.004 kgCharge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]Final velocity of bead = 3 m/sTo determine the magnitude of the potential difference (V):
How to calculate the potential difference (V).We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:
[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]
V = 900 V.
Read more on kinetic energy here: https://brainly.com/question/1242059
is a step in the scientific method. The step that follows this step involves forming
Answers
Answer: read this hope this helped
Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these
Answers
Answer:
a) E = 0
b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
From which we have;
[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]
E = 0/A = 0
E = 0
b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]
[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]
By Gauss theorem, we have;
[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]
Therefore, we get;
[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]
The electrical field outside the spherical shell
[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]
[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]
Therefore, we have;
[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential
Answers
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.
Solution:
The relation between electric field and the electric potential is:
E = [tex]\frac{dV}{dr}[/tex]
So, making dV the subject, we have:
dV = E x dr
Integrating the above equation, we get.
V = [tex]\int\limits^_ {} \,[/tex]E x dr Equation 1
Now, in 2-D
E is inversely proportional to the radius r.
E ∝ 1/r
So, we can write: replacing E ∝ 1/r in the equation 1
V ∝ [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr
Which implies that,
V ∝ log (r)
Hence, distance dependence expected for the electric potential = ln (r)