what is the mathematical definition of momentum? what is a more conceptual or descriptive definition of momentum?
Answer:
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion.
Explanation:
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
Question 18 of 25
Which type of reaction is shown in this energy diagram?
Energy
Products
Activation
Energy
Reoctants
to
ti
Time
A. Endothermic, because the products are lower in energy
B. Exothermic, because the reactants are lower in energy
C. Endothermic, because the reactants are lower in energy
D. Exothermic, because the products are lower in energy
Answer:
Endothermic, because the reactants are lower in energy (C)
Explanation:
From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that when the enthalpy change Eproducts is gretater than Ereactants, the reaction is said to be endothermic.
A heat pump is used to heat a building. The external temperature is lower than the internal temperature. The pump's coefficient of performance is 3.70, and the heat pump delivers 7.27 MJ as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it
Answer:
Heat pump needs 1.965 megajoules each hour to run.
Explanation:
The Coefficient of Performance ([tex]COP[/tex]), no unit, of a Carnot's heat pump is:
[tex]COP = \frac{Q_{H}}{W}[/tex] (1)
Where:
[tex]Q_{H}[/tex] - Heat received by the building, measured in megajoules.
[tex]W[/tex] - Work needed to run the heat pump, measured in megajoules.
If heat pump is a Carnot engine working in reverse, then the amount of work needed to run the heat pump is the least possible work. If we know that [tex]Q_{H} = 7.27\,MJ[/tex] and [tex]COP = 3.70[/tex], then the amount needed by the heat pump each hour is:
[tex]W = \frac{Q_{H}}{COP}[/tex]
[tex]W = \frac{7.27\,MJ}{3.70}[/tex]
[tex]W = 1.965\,MJ[/tex]
Heat pump needs 1.965 megajoules each hour to run.
To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal length of the eyepiece lens should be ________. To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal length of the eyepiece lens should be ________. small; small small; large large; small large; large
Answer:
Large; small.
Explanation:
A telescope can be defined as an optical instrument or device which comprises of a curved mirror and lenses used for viewing distant objects i.e objects that are very far away such as stars and other planetary bodies. The first telescope was invented by Sir Isaac Newton.
To have the highest magnification in a telescope, the focal length of the objective lens should be large and the focal length of the eyepiece lens should be small.
This ultimately implies that, the eyepiece lens has a small focal length while the objective lens has a large focal length.
What composes about 71% of Earth's outermost layer?
A
oceanic crust
B
asthenosphere
С
lithosphere
D
continental crust
Answer:
A. oceanic crust
Explanation:
I remember that the ocean is said to cover 71% of the Earth's surface. If you look at a globe, notice that most all the surface is blue like the ocean.
One other note: the surface is the Earth's outermost layer. Think of it this way: surface implies the top of something, something exposed to the outside.
Therefore, the answer is A. Hope this helps you understand the question more! Have a great day, 'kay?
If the coefficient of kinetic friction is 0.43 for a box sliding across your lab table and the
box weighs 7.4 N, what is the force of kinetic friction?
Answer:
3.2N
Explanation:
Given parameters:
Coefficient of kinetic friction = 0.43
Weight of box = 7.4N
Unknown:
Force of kinetic friction = ?
Solution:
The force of kinetic friction is given as:
Force of kinetic friction = UN
U is the coefficient of friction
N is the weight
Force of kinetic friction = 0.43 x 7.4 = 3.2N
a long solid rod 4.5 cm in radius carries a uniform volume charge density. if the electric field strength at the surface of the rod (not near either end) is 16 kn/c, what is the volume charge density
Answer:
6.29 μC/m³
Explanation:
Volume charge density is the quantity of charge per unit volume.
The direction of the electric field was not specified, therefore the volume charge density (ρ) is given by:
2πRLE = ρπR²L/ε₀
ρ = 2Eε₀ / R
Where E = electric field strength = 16 kN/C = 16 * 10³ N/C, R = radius of rod = 4.5 cm = 0.045 m, ε₀ = relative permittivity of free space = 8.85 * 10⁻¹² C² / Nm²
Therefore:
ρ = 2(16 * 10³ N/C)(8.85 * 10⁻¹² C²/Nm²) / 0.045 m = 6.29 * 10⁻⁶ C/m³
ρ = 6.29 μC/m³
Tasks
Task 2 - Compare and contrast the use of D'Alembert's principle with
the principle of conservation of energy to solve an
engineering problem
A motor vehicle having a mass of 800 kg is at rest on an incline of 1 in 8 when the
brakes are released. The vehicle travels 30 m down the incline against a constant
frictional resistance to motion of 100 N where it reaches the bottom of the slope.
a) Using the principle of conservation of energy, calculate the velocity of the
vehicle at the bottom of the incline.
b) Using an alternative method that does not involve a consideration of energy,
cacluate the velocity of the vehicle at the bottom of the incline.
c) Discuss the merits of the two methods you have used for parts a) and b) of
this question. Justify the use of an energy method for these types of
problems.
Answer:
NE DIYON INGILIZ MISIN SEN
Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel (stirrer) that transfers energy to the air at a constant rate of 10 W for 1h. If no changes in kinetic or potential energy occur, determine a) The specific volume at the final state, in m3 /kg. b) The energy transfer by work, in kJ. c) The change in specific internal energy of the air, in kJ/kg.
Answer:
[tex]0.3\ \text{m}^3/\text{kg}[/tex]
[tex]36\ \text{kJ}[/tex]
[tex]18\ \text{kJ/kg}[/tex]
Explanation:
V = Volume of air = [tex]0.6\ \text{m}^3[/tex]
P = Power = 10 W
t = Time = 1 hour
m = Mass of air = 2 kg
Specific volume is given by
[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]
The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]
Work done is given by
[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]
The energy transfer by work, is [tex]36\ \text{kJ}[/tex]
Change in specific internal energy is given by
[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]
The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]
HELP ASAP PLS
A balloon with a positive charge will stick to a wall that has a negative charge.
What force causes this?
A. Gravity
B. Electric force
C. Magnetic force
D. Air gesistance
Did I hear correctly that the speed of light is different in deep space observation?
Answer:
Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.
The driver of a 3000 lb. car, coasting down a hill, sees a red light at the bottom, and must stop. His speed when he applies the brakes is 60 mph, and he is 100 feet (vertically) above the bottom of the hill. (a)How much energy as heat must be dissipated by the brakes if we neglect wind resistance and other frictional effects
Answer:
Explanation:
60 mph = 60 x 1760 x 3 / (60 x 60) ft /s
speed of car , v = 88 ft /s
kinetic energy of car = 1/2 m v²
= .5 x 3000 x 88²
= 11616 x 10³ poundal - foot
Potential energy = mgh
= 3000 x 32 x 100
= 9600 x 10³ poundal - foot
Total energy = potential energy + kinetic energy
= ( 11616 + 9600 )x 10³
= 21216 x 10³ poundal - foot .
This energy is dissipated as heat when brakes are applied on the car to stop the car .
The superheroine Xanaxa, who has a mass of 65.1 kg , is pursuing the 78.7 kg archvillain Lexlax. She leaps from the ground to the top of a 153 m high building then dives off it and comes to rest at the bottom of a 17.5 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy
Answer:
There is net loss of gravitational energy .
Explanation:
When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .
Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .
Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
Answer:
the Potential Energy is 2.304 × 10⁻¹⁶ J
Explanation:
Given the data in the data in the question;
The expression for the electric potential energy between the charges can be expressed as follows;
PE = qV ------equ 1
where q is the charge and V is the electric potential
Also the formula for electric potential due to point a point in a field is;
V = kq / r -------equ 2
where k is the electrostatic constant and r is the distance form the charged particle
input equation 2 into 1
PE = q × kq / r
PE = kq²/r ------- equ 3
so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )
PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²
PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²
PE = 2.304 × 10⁻¹⁶ J
Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J
An electric charge at rest produce
Answer:
Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.
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It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil would it take to make a capacitorlarge enoughto hold this amount of energy if one were to use plastic garbage bag with a 2.6 x 10-5m thickness that breaks down at 610 volts as a dielectric
Answer:
L = 1.11 x [tex]10^{6}[/tex] m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x [tex]10^{5}[/tex] m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x [tex]10^{-2}[/tex] m
Length = ?
So,
we know that,
U = 1/2 C Δ[tex]v^{2}[/tex]
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ[tex]v^{2}[/tex]
52000 J = (0.5) x (C) x ([tex]610^{2}[/tex])
C = 0.28 F
And we also know that,
C = [tex]\frac{K*E*A}{d}[/tex]
E = 8.85 x [tex]10 ^{-12}[/tex]
K = 3.7
A = 0.20 x L
d = 2.6 x [tex]10^{5}[/tex] m
Plugging in the values into the formula, we get:
0.28 = [tex]\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }[/tex]
Solving for L, we get:
L = 1.11 x [tex]10^{6}[/tex] m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1 m/s2 for 4 seconds. It then continues at a constant speed for 12.9 seconds, before getting tired and slowing down with constant acceleration coming to rest 66 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. 1)How fast is the hare going 1.6 seconds after it starts
Answer:
v = 4 m/s
Explanation:
Given that,
Initial speed of hare, u = 0
The hare accelerates uniformly at a rate of 1 m/s² for 4 seconds.
We need to find how fast is the hare going 1.6 seconds after it starts. Let the speed be v. So,
v = u+at
Substitute all the values,
v = 0+1×4
v = 4 m/s
So, the required speed of the hare is 4 m/s after it starts.
A pendulum has a period of 5.14s and a length of 0.25m. What is the acceleration
due to gravity? *
Answer:
Acceleration due to gravity, g = 2.68m/s²
Explanation:
Given the following data;
Period = 5.14s
Length = 0.25m
To find acceleration due to gravity, g;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]
[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]
[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]
[tex] 0.8180 = \sqrt {0.25g} [/tex]
Taking the square of both sides
[tex] 0.8180^{2} = 0.25g [/tex]
[tex] 0.6691 = 0.25*g[/tex]
[tex] g = \frac {0.6691}{0.25} [/tex]
Acceleration due to gravity, g = 2.68m/s²
Which formula is used to find an objects acceleration
Answer:
a=∆v/∆t
Explanation:
The definition of Acceleration is the change in velocity in a given time. So this means you first calculate ∆v (Change in velocity), and you calculate ∆t which is the time taken to apply that change in velocity. Then you find a= ∆v/∆t. This gives us the equation of Acceleration.
Answer:
C. a=∆v/∆tExplanation:
5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
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The attached picture shows Bernoulli's Theorem.
Given a volume of 1000. Cm^3 of an ideal gas at 300 k, what volume would iy occupy at a temperature of 600 k
Answer:2000 cm³
Here, pressure remains constant.
So, b the gas law
V/V' = T/ T'
1000 / V' = 300 / 600
V' = 2000 cm³
Explanation:also pls mark brainliest
Which object would have the greatest acceleration?
Answer:
D
Explanation:
A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.
A major source of water pollution comes from that washes chemicals and other pollutants from improperty managed land.
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
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Two balls are thrown against a wall with the same velocity. The first ball is made of rubber and bounces straight back with some non-zero speed. The second ball is made of clay and sticks to the wall after impact. If we assume the collision time was the same for each ball, which ball experienced a greater average acceleration during the collision with the wall? A. the average acceleration was the sameB. the clay ball C. there is not enough information D. the rubber ball
Answer:
A. the average acceleration was the same
Explanation:
Acceleration is calculated by finding the difference of the initial velocity from the final velocity (on impact, usually 0) and then dividing by the amount of time that took place. If we assume that both balls were thrown at the same initial force, and ended up hitting the wall at the same time then we can say that the average acceleration was the same. If the initial velocity was not the same then we would need the initial velocity of each ball in order to calculate the acceleration of each object and determine which had a greater acceleration.
A car has a mass of 2000 kg and accelerates at 2 meters per second per second. What is the magnitude of the net force exerted on the car?
Hello!
[tex]\large\boxed{4000 N}[/tex]
Use the following equation to solve for the net force (N):
∑F = m × a
Plug in the given mass (kg) and speed (m)
∑F = 2000 * 2
Simplify:
∑F = 4000 N
Help me out please. It’d be greatly appreciated
Answer:
Option D
2Na + Cl₂ —> 2NaCl
Explanation:
We'll begin by stating the law of conservation of matter.
The law of conservation of matter states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.
For an equation to comply with the law of conservation of matter, the number of atoms of each element must be the same on both side of the equation. This simply means that the equation must be balanced!
NOTE: An unbalanced equation simply means matter has been created or destroyed.
Now, we shall determine which equation is balanced. This can be obtained as follow:
For Option A
Na + Cl₂ —> 2NaCl
Reactant:
Na = 1
Cl = 2
Product:
Na = 2
Cl = 2
Thus, the equation is not balanced!
For Option B
2Na + 2Cl₂ —> 2NaCl
Reactant:
Na = 2
Cl = 4
Product:
Na = 2
Cl = 2
Thus, the equation is not balanced!
For Option C
2Na + Cl₂ —> NaCl
Reactant:
Na = 2
Cl = 2
Product:
Na = 1
Cl = 1
Thus, the equation is not balanced!
For Option D
2Na + Cl₂ —> 2NaCl
Reactant:
Na = 2
Cl = 2
Product:
Na = 2
Cl = 2
Thus, the equation is balanced!
From the above illustrations, only option D has a balanced equation. Thus, option D illustrate the law of conservation of matter.
Which is a valid velocity reading for an object?
45 m/s
45 m/s north
O 0 m/s south
O 0 m/s
Answer:45 m/s north
Explanation:
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
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