A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 0.00820 m. A beam of alpha particles (charge +2e, mass 6.64Ã10â27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field.What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?
Answers
Answer:
B = 4.45mT in the +^k direction
Explanation:
In order to calculate the required magnitude of the magnetic force, to achieve that the beam of particles emerge undeflected of the parallel plates, the electric force between the plates and the magnetic field in that region must be equal.
[tex]F_E=F_B\\\\qE=qvB[/tex] (1)
q: charge of the particles beam = +2e = 2*1.6*10^-19C
v: speed of the particles = ?
B: magnitude of the magnetic field = ?
E: electric field between the plates = V/d
V: potential difference between the parallel plates = 150V
d: distance of separation of the plates = 0.00820m
If you assume that the below plate is negative, the electric force on the particles has a direction upward (+^j). Then, the direction of the magnetic force must be downwards (-^j).
To obtain a downward magnetic force, it is necessary that the magnetic field point out of the page. In fact, if the direction of motion of the particles is toward east (+^i) and the magnetic field points out of the page (+^k), you have:
^i X ^k = -^j
Furthermore, it is necessary to calculate the sped of the particles. It is made by using the information about the charge, the potential difference that accelerates the particles and the kinetic energy.
[tex]K=qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}[/tex] (2)
You replace the expression (2) into the equation (1) and solve for B:
[tex]B=\frac{E}{v}=E\sqrt{\frac{m}{2qV}}[/tex]
[tex]B=\frac{V}{d}\sqrt{\frac{m}{2qV}}\\\\B=\frac{150V}{0.0820m}\sqrt{\frac{6.64*10^{-27}kg}{2(2(1,6*10^{-19}C))(1.75*10^3V)}}\\\\B=4.45*10^{-3}T=4.45mT[/tex]
The required magnitude of the magnetic field is 4.45mT and has a direction out of the page +^k
Following are the solution to the given question:
In order to emerge using reflecting means, use the following formula:
[tex]\to F_E = F_B ..............(1)\\\\ \to F_E = \text{electric force}\\\\ \to F_B = \text{magnetic force}\\\\[/tex]
Calculating the Lorent's force:
[tex]\to F=qE+qv \times B \ \ also,\ \ K_{E} =U_{E} \\\\[/tex]
[tex]\to K_{E}[/tex][tex]= \text{kinetic energy} = -\frac{1}{2} \ mv^2 \\\\[/tex]
[tex]\to U_{E} = \text{potential energy} = q_V[/tex]
Calculating the value of v: \\\\
[tex]\to v= \sqrt{\frac{2qV}{m}} \\\\ \to q = 2e^{+} = 2 (1.6 \times 10^{-19}\ C) = 3.2 \times 10^{-19} C \\\\\to V = 1.75 \times 10^{3} \V \\\\\to m = 6.64 \times 10^{-27} \ kg\\\\ \to v = 410,700.25 \ \frac{m}{s}\\\\[/tex]
It's the particle's velocity, but the velocity also is defined as:
[tex]\to v=\frac{E}{B}[/tex]
solving for B:
[tex]\to B= \frac{E}{v}\\\\[/tex]
[tex]= \frac{\frac{V}{d}}{v}\\\\ =\frac{V}{vd} \\\\= \frac{150\ V}{(410,700.25 \ \frac{m}{s}) (8.2 \times 10^{-3} m)} \\\\= 0.045\ T\\\\[/tex]
When indicated in the diagram, the direction is parallel to "v" and E.
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Related Questions
What is the equivalent resistance between the points A and B of the network?
Answers
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Answers
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.
Answers
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.00m/s. Determine the constants A and B with units.
Answers
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
g Suppose that you charge a 3 F capacitor in a circuit containing eight 3.0 V batteries, so the final potential difference across the plates is 24.0 V. How much charge is on each plate
Answers
The complex, highly technical formula for capacitors is
Q = C V
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
Charge = 72 Coulombs
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average lightning bolt, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around 15 coulombs of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal
Answers
Answer:
Halide
Explanation:
It has at least one element from the halogen group (17)
Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.
What are halides?When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.
These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.
Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.
Thus, halides are compounds that have at least one element from group 17.
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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis
Answers
Answer:
M1 Vx1 + M2 Vx2 + M3 Vx3 = 0 conservation of momentum in x direction
Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3
Vx3 = - 320 * 2 / 100 = -6.4 m/s M2 has no x-component of momentum
Likewise:
Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3
Vy3 = - 355 * 1.5 / 100 = -5.33 m/s
tan theta = -5.33 / -6.4 = .833 where theta is in the third quadrant and measured from the negative x-axis
theta = 39.8 deg
180 + 39.8 = 219.8 from the positive x-axis
how is mirage formed
Answers
Answer:
Mirages are formed when the ground is really hot and the air is really cold. The hot ground will warm a layer of the air closest to the ground. When the light moves through the cold air and hits the warm air it bends creating the U shaped bend
Explanation:
Hope this helps!
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answers
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
According to the law of conservation of energy, if the ocean water cools, then something else should warm. What is it that warms?
Answers
Answer:
The answer is air
Explanation:
Suppose that the voltage of the battery in the circuit is 3.9 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the paper) is 1.1 T, and the length of the rod between the rails is 0.22 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answers
Answer:
v = 16.11 m/s
Explanation:
In order to calculate the maximum speed of the rod, you use the following formula:
[tex]\epsilon=vBLsin\theta[/tex] (1)
ε = voltage of the circuit = 3.9V
v: maximum speed of the rod = ?
B: magnitude of the magnetic field = 1.1T
L: length of the rod = 0.22m
θ: angle between the direction of motion of the rod and the direction of the magnetic field = 90°
You solve the equation (1) for v and replace the values of the other parameters:
[tex]v=\frac{\epsilon}{BLsin\theta}=\frac{3.9V}{(1.1T)(0.22m)sin90\°}\\\\v=16.11\frac{m}{s}[/tex]
The maximum speed of the rod is 16.11 m/s
B. Write short notes on:
1. Horticulture
2. Pisciculture
3. Aviculture
4. Veterinary science
5. Intensive farming.
Answers
1. Horticulture is the agriculture of plants, mainly for food, materials, comfort and beauty for decoration.
2.Pisciculture also known as fish farming is the rearing of fish for food in enclosures such as fish ponds or tanks.
3.Aviculture is the practice of keeping and breeding birds, especially of wild birds in captivity. Aviculture is generally focused on not only the raising and breeding of birds, but also on preserving avian habitat, and public awareness campaigns.
4. Veterinary medicine is the branch of medicine that deals with the prevention, control, diagnosis, and treatment of disease, disorder, and injury in animals. Along with this, it also deals with animal rearing, husbandry, breeding, research on nutrition and product development.
5. Intensive agriculture, also known as intensive farming and industrial agriculture, is a type of agriculture, both of crop plants and of animals, with higher levels of input and output per cubic unit of agricultural land area.
Hope this helps.
A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?
Answers
Answer:
16,931 turnsExplanation:
The magnetic field produced is expressed using the formula
[tex]B = \frac{\mu_0NI}{L}[/tex]
B is the magnetic field = 0.30T
I is the current produced in the coil = 4.5A
[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A
L is the length of the solenoid = 32 cm = 0.32 m
N is the number of turns in the solenoid.
Making N the subject of the formula from the equation above;
[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]
[tex]N = \frac{BL}{\mu_0I}[/tex]
Substituting the give values to get N;
[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]
The number of turns the solenoid must have is approximately 16,931 turns
A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answers
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
How much work is needed to move an object from one position to another when both positions are located the same distance from the center of the earth
Answers
Answer:
The product of the object's weight and the horizontal distance between the two positions.
Explanation:
Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answers
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Answers
Answer:
The change in gravitational potential energy of the climber-Earth system is [tex]\Delta PE = 396900 \ J[/tex]
Explanation:
From the question we are told that
The mass of the hiker is [tex]m = 75 \ kg[/tex]
The time taken is [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]
The vertical elevation after time T is [tex]H = 540 \ m[/tex]
The change in gravitational potential is mathematically represented as
[tex]\Delta PE = mgH[/tex]
here g is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
substituting values
[tex]\Delta PE = 75 * 9.8 * 540[/tex]
[tex]\Delta PE = 396900 \ J[/tex]
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answers
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
the time required for one cycle, a complete motion that returns to its starting point, it called the_____. period medium frequency periodic motion
Answers
Answer:
The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion
Explanation:
I hope this will help you:)
Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.Required:Which rock has the greatest speed just before it hits the ground? Assume air resistance is negligible.
Answers
Answer:Three rocks of equal mass are thrown
Explanation:
The phenomena: Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.
"A plane has an airspeed of 142 m/s. A 16.0 m/s wind is blowing southward at the same time as the plane is flying. If the velocity of the plane relative to Earth is due east, what is the magnitude of that velocity
Answers
Answer:
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
Explanation:
Given;
A plane has an airspeed of 142 m/s (eastward)
vi = 142 m/s
16.0 m/s wind is blowing southward at the same time as the plane is flying
vb = 16.0m/s
Writing the relative velocity vector, we have;
Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.
v = 142i - 16j
The magnitude of the velocity is;
vr = √(vi^2 + vb^2)
vr = √(142^2 + 16^2)
vr = √(20420)
vr = 142.8985654231 m/s
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass m1 attached to the center of the rod and a particle of mass m2 attached to the outer end of the rod. The inner section of the rod sustains a tension that is 5 times as great as the tension that the outer section sustains. Find the ratio m1/m2
Answers
Answer:
m₁/m₂ =8
Explanation:
Writing the centripetal force equation: where T₁ is inner tension and T₂ is outer tension
T₂ = m₂ × ω²R....equation 1
T₁-T₂ = m₁ × ω²R/2
Also T₁=5T₂
5T₂-T₂ = m₁ × ω²R/2
4 × 2T₂ = m₁ × ω²R
8T₂ = m₁ × ω²R
dividing it by equation 1
8 = m₁/m₂
Hence m₁/m₂ =8
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A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answers
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
Explain the purpose of hot gravity filtration. Why is it good to use the stemless funnel for this experiment
Answers
Answer: It is done to prevent the necessary compound from solidifying along with the debasements. It expels any insoluble pollutions from the appropriate response (as opposed to separating the predetermined item). With since quite a while ago stemmed channels, the gems kick off inside the progression because the arrangement cools, obstructing the pipe. utilizing a stemless channel keeps this from occurring.
Explanation:
it is good to use the stemless funnel for hot gravity filtration experiment, to prevent the necessary compound from solidifying, expels any insoluble pollutions from the appropriate response.
what is hot gravity filtration ?Recrystallization is the process of getting pure crystals from an impure compound in a solvent and Hot gravity filtration remove the impurities from a solution prior to recrystallization.
In this technique the filtration equipment and the sample are heated and the filtration is needed for recrystallization which requires a hot solution as it need to be supersaturated for crystals to form on cooling.
Hot solutions hold more solute in a suspension than a cold solution as the solubility of solids increases with a increase in temperature, that means saturated solution contain more dissolved solute.
When the hot solution cool down, it will be supersaturated and hold more dissolved solute than its cold. The main objective to choose a solvent is that it dissolves the compound when heated, but that doesn’t dissolve the impurity at high temperatures.
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During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
1) = what work do they do?
2) = what power do they develop if the work is done in 5.5 minutes?
Answers
Answer:
work is =7590joules
power = 23watts
Answer:
1) 7590 Joules
2) 23 Watts
Explanation:
1) Work = force × distance
W = mgh
W = (115 kg) (10 m/s²) (6.6 m)
W = 7590 J
2) Power = work / time
P = W / t
P = (7590 J) / (330 s)
P = 23 W
What is the equivalent temperature in kelvin if you have a metal at 50°F?
Answers
Answer:
The required temperature is 283 K.
Explanation:
[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]
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Applying Gaussâs Law
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 μC charge put on a 5.0-cm aluminum spherical ball at the following two points in space: (a) a point 1.0 cm from the center of the ball (an inside point) and (b) a point 10 cm from the center of the ball (an outside point).
Answers
Answer:
a) E = 0
b) E = 2.697 MN/C
Explanation:
Solution:-
- The Gauss Law makes life simpler by allowing us to determine the Electric Field strength ( E ) of symmetrically charged objects. By choosing an appropriate Gaussian surface and determine the flux ( Φ ) that passes through an imaginary closed surface.
- The Law states that the net flux ( Φ ) that passes through a Gaussian surface is proportional to the net charged ( Q ) stored within that surface. We can mathematically express the flux ( Φ ) as follows:
Φ = Q / εo
Where, 1 / εo : The proportionality constant
εo: The permittivity of free space = 8.85*10^-12
- The flux produced by a charged object is also given in form of a surface integral of Electric Field ( E ) over the entire surface area ( A ) of the Gaussian surface as follows:
Φ = [tex]_S\int\int [ E ] . dA[/tex]
- We can combine the two relations as follows:
[tex]_S\int\int [ E ] . dA[/tex] = Q / εo
- Now we will consider a charged metal sphere. The important part to note is that the charge on a conducting sphere ( Q ) uniformly distributed on the outside surface of the charged sphere.
- Lets consider a case, where we set up our Gaussian surface ( spherical ) with radius ( r ) < radius of the charged metal surface ( a ). We will use the combined relation and determine the Electric Field ( E ) within a charged metal sphere as follows:
[tex]E. ( 4\pi*r^2 ) = \frac{Q_e_n_c}{e_o} \\\\E = \frac{Q_e_n_c}{e_o4\pi*r^2}[/tex]
- However, the amount of charge enclosed in our Gaussian surface is null or zero. As all the charge is on the surface r = a. Hence (Q_enc = 0 ),
[tex]E = 0[/tex] ..... ( r < a )
- For the case when we set up our gaussian surface with radius ( r ) > radius of the charged metal surface ( a ). We placed a charge of Q = +3.0uC on the surface of the metal sphere. Therefore, the electric field strength at a distance ( r ) from the center of metal sphere is:
[tex]E = \frac{Q_e_n_c}{e_o*4*\pi*r^2 } = k\frac{Q_e_n_c}{r^2 }[/tex] .... ( r > a )
- The above relation turns out to be the Electric Field strength ( E ) produced by a point charge at distance ( r ) from the center. Where, k = 8.99*10^9 is the Coulomb's constant.
a) The radius of the charged metal sphere is given to be a = 5.0 cm. The first point r = 1.0 cm lies within the metal sphere. We looked at the first case where, ( r < a ) the enclosed charge is zero. Hence, the magnitudue of Electric Field Strength ( E ) is zero. ( E = 0 )
b) The second point lies at 10 cm from the center. For this we will use the second case where, ( r > a ). The Electric Field Strength due to a point charge with an enclosed charge of Q = +3.0 uC is:
[tex]E = ( 8.99*10^9 ) * \frac{3.0*10^-^6}{0.1^2} \\\\E = 2697000 N / C[/tex]
Answer: The electric field strength at point 10 cm away from the center is 2.697 MN/C
Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predominate when the X-rays hit the metal?
a) Photoelectric Effect 3
b) Compton Effect 3
c) Pair Production
Answers
Answer:
the answer is option A = photoelectric effect
Explanation:
If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.
The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) When 2.0 mol of O2 are consumed in this reaction, ________ mol of CO2 are produced.
Answers
Answer:
1.2
Explanation:
2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂
5. Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car
Answers
Answer:
The work done by both Joel and Jerry is equal to 0 J.
Explanation:
The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,
W = F.d
where,
W = Work Done on the Body
F = Force Applied on the Body
d = displacement covered by the body
In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,
W = F(0)
W = 0 J (For both Joel and Jerry)