Answer:
yes it would.
Explanation:
Which of the following best describes what an electron configuration is?
A) An electron configuration is the mathematical symbol used to describe the first three principle energy levels in which electrons are found within an atom.
B) An electron configuration is the system used to name each principle energy level and the sublevels in which electrons are found within an atom.
C) An electron configuration is used to name the sublevels in which electrons are found within an atom.
D) An electron configuration describes how the principle energy level and sublevels in which protons and neutrons are found affect the electrons within an atom.
Answer:
An electron configuration is used to name the sublevels in which electrons are found within an atom
Explanation:
An electron configuration is used to name the sublevels in which electrons are found within an atom for example for the atom carbon has 6electrons so its configuration is 1s²2s²2p²
With the 1s 2s and 2p showing the various energy sublevels
Concerning the concepts and fundamentals of the atomic model, an electron configuration is used to name the sublevels in which electrons are found within an atom. Hence, option (C) is correct.
What is electronic configuration?The arrangement of electrons in specific energy levels around the atomic nucleus of an atom is known as the electron configuration of an atom.
As per the atomic shell model, "The electrons occupy the various energy levels from the first shell nearest to the nucleus, upto the farthest shell from the nucleus.
Consequently, an electron configuration is used to name the sublevels in which electrons are found within an atom. For example for the atom carbon has 6 electrons so its configuration is 1s²2s²2p². With the 1s 2s and 2p showing the various energy sublevels.
Thus, we can conclude that an electron configuration is used to name the sublevels in which electrons are found within an atom.
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Why does pressure increases with depth?
Answer:
Explanation:
This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .
which surface has more friction
A. an ice rink
B. a grassy field
C. a paved road
Answer:
The answer is C. a paved roadExplanation:
Generally, roads are meant to have very rough surfaces to allow for friction between car tyres hence the motion of vehicles.
What is a paved surface?
Paved roads are roads that are covered with a firm surface suitable for travel, like paving stones or concrete or asphalts.
Another good example of a paved road is a small paved courtyard, covered with a hard layer of concrete/cement.
The other two examples sited in this problem are glossy materials with very low surface friction, that is their surfaces are very smooth.
A solid sphere of radius R has a uniform charge density and total charge Q. Find the total energy of the sphere U in terms of
Answer:
[tex]\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}[/tex]
Explanation:
From the information given;
The surface area of a sphere = [tex]4 \pi r ^2[/tex]
If the sphere is from the collection of spherical shells of infinitesimal thickness = dr
Then,
the volume of the thickness and the sphere is;
V = [tex]4 \pi r ^2 \ dr[/tex]
Using Gauss Law
[tex]V(r) = \dfrac{k_cq(r)}{r}[/tex]
here,
q(r) =charge built up contained in radius r
since we are talking about collections of spherical shells, to work required for the next spherical shell r +dr is
[tex]-dW= dU = V(r) dq = \dfrac{k_c \ q(r)}{r} \ dq[/tex]
where;
[tex]q (r) = \dfrac{4}{3} \pi r^3 \rho[/tex]
dq which is the charge contained in the next shell of charge
here dq = volume of the shell multiply by the density
[tex]dq = 4 \pi r^2 \ dr \ \rho[/tex]
equating it all together
[tex]dU = \dfrac{k_c \frac{4}{3} \pi r^3 \rho}{r} 4 \pi r^2 \ dr \ \rho = \dfrac{16 \pi^2 \ k_c \ \rho^2}{3} \ r^4 \ dr[/tex]
Integration the work required from the initial radius r to the final radius R, we get;
[tex]U = \int^R_0 \ dU[/tex]
[tex]U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} r^4 \ dr[/tex]
[tex]U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} [\dfrac{r^5}{5}]^R_0[/tex]
[tex]U = \dfrac{16 \pi^2 k_c \rho^2}{15} \ R^5[/tex]
Recall that:
the total charge on a sphere, i.e [tex]Q_{total} = \dfrac{4}{3} \pi R^3 \rho[/tex]
Then :
[tex]\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}[/tex]
g a long, straight wire carries a current an electron Is it necessary to include the effects of gravity?
Answer:
The force is [tex]3.2\times10^{-20}\ N[/tex]
Explanation:
Given that,
A long, straight wire carries a current.
Suppose, a long, straight wire carries a current 0.86 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling at speed of [tex]6\times10^{4}\ m/s[/tex] directly toward the wire. What are the magnitude of the force that the magnetic field of the current exerts on the electron?
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}I}{4\pi r}[/tex]
Where, I = current
r = distance
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times0.86}{2\pi\times4.5\times10^{-2}}[/tex]
[tex]B=0.0000038\ T[/tex]
[tex]B=3.8\times10^{-6}\ T[/tex]
We need to calculate the force
Using formula of force
[tex]F=qvB\sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
Where, q = charge of electron
v = velocity of electron
B = magnetic field
Put the value into the formula
[tex]F=1.6\times10^{-19}\times6\times10^{4}\times3.8\times10^{-6}\sin90[/tex]
[tex]F= 3.2\times10^{-20}\ N[/tex]
Hence, The force is [tex]3.2\times10^{-20}\ N[/tex]
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
This question is incomplete
Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 2.8 s .
Answer:
2.15 mV
Explanation:
The complete question is
You have a coil of wire with 17 turns each of 1.5 cm radius. You place the plane of the coil perpendicular to a 0.50-TB? field produced by the poles of an electromagnet.
Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 1.9 s .
Radius of the coil = 1.5 cm = 0.015 m
number of turns = 17
Initial magnetic field [tex]B_{1}[/tex] = 0.50 T
Final magnetic field [tex]B_{2}[/tex] = 0 T
time taken dt = 1.9 s
Area pf the coil = π[tex]r^{2}[/tex] = 3.142 x [tex]0.015^{2}[/tex] = 7.1 x 10^-4 m^2
magnetic flux = BA
initial flux = [tex]B_{1}[/tex]A = 0.5 x 7.1 x 10^-4 = 3.55 x 10^-4 Wb
Final flux = [tex]B_{2}[/tex]A = 0 x 7.1 x 10^-4 = 0 Wb
change in flux dФ = 3.55 x 10^-4 - 0 = 3.55 x 10^-4 Wb
Induced EMF E = NdФ/dt = (17 x 3.55 x 10^-4)/2.8 = 2.15 x 10^-3 V
==> 2.15 mV
At what speed, in m/s, would a moving clock lose 4.5 ns in 1.0 day according to experimenters on the ground
Answer:
the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
Explanation:
Let's first find the relation for the time dilation by using the formula:
[tex]\Delta t' = \Delta t \gamma[/tex]
Making [tex]\gamma[/tex] the subject of the formula, we have:
[tex]\gamma= \dfrac{\Delta t'}{\Delta t}[/tex]
Number of day(s) = 1
1 day in seconds = 24 × 60 × 60
= 84600 seconds
[tex]\gamma= \dfrac{86400 \ s}{86400 \ s - 4.5 \ ns}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= \dfrac{86400 \ s - 4.5 \ ns}{86400 \ s}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= 1 - 5.4 \times 10^{-14}[/tex]
From the above equation, if we apply binomial expansion to it, we have the following,
[tex]\sqrt{1- \dfrac{v^2}{c^2} }=(1-\dfrac{v^2}{c^2})^{\frac{1}{2}}[/tex]
⇒[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2})[/tex]
So;
[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 1 - 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 2 \times 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 1.08 \times 10^{-13}[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times c^2[/tex]
where ;
[tex]c = 3 \times 10^8 \ m/s[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2}[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} } \times (3 \times 10^8 \ m/s)[/tex]
[tex]v =3.286 \times 10^{-7} \times (3 \times 10^8 \ m/s)[/tex]
[tex]\mathbf{v =98.58 \ m/s}[/tex]
Therefore, the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
What is the binding energy (in MeV per nucleon) for the ⁶₃Li nucleus?
Answer:
Explanation:
⁶₃Li will have 3 protons and 3 neutrons .
mass of proton in amu = 1.00727 amu
mass of neutron in amu = 1.00866 amu
mass of lithium nucleus in amu = 6.01512 amu
mass defect = 3 ( 1.00727 + 1.00866 ) - 6.01512 amu
= .03267 amu
Binding energy = mass defect in amu x 931 Mev
= 30.41 MeV
binding energy per nucleon
no of nucleon = 3 + 3 = 6
binding energy per nucleon = 30.41 / 6 Mev
= 5.068 MeV .
The binding energy of ⁶₃Li nucleus in MeV per nucleon is 5.26 MeV per nucleon.
We can see that Li has three protons and three neutrons. The total mass of the Li nucleus is obtained from;
3(1.007277) + 3(1.008665) = 3.023 + 3.026 = 6.049 amu
Actual mass of Li - 6 = 6.0151 amu
Mass defect = 6.049 amu - 6.0151 amu = 0.0339 amu
We can obtain the binding energy in MeV as follows;
Binding energy = 0.0339 amu × 931 = 31.56 MeV
The binding energy per nucleon in MeV per nucleon = 31.56 MeV/6
= 5.26 MeV per nucleon
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The prediction that alcohol slows reaction time is:_________.
a) nondirectional
b) directional
c) semi directional
d) partially directional
Answer:
b) directional
Explanation:
The prediction that alcohol slows reaction time is Directional . This is because for alcohol to be known as one which slows reaction time means there have been various hypothesis conducted which supports this.
In this case, formulation of a null hypothesis is usually necessary which means that alcohol does not slow reaction time and another alternative hypothesis that suggests that alcohol slows reaction time.
The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 2.45 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
Answer:
1.45
Explanation:
If the position versus time graph of an object is a horizontal line, the object is
In a position-time graph, the velocity of the moving object is represented by the slope, or steepness, of the graph line. If the graph line is horizontal, like the line after time = 5 seconds in Graph 2 in the Figure below, then the slope is zero and so is the velocity. The position of the object is not changing.
Explanation:
The slope of a position time graph gives the velocity of an object. It can be calculated as follows:
[tex]v=\dfrac{dx}{dt}[/tex]
Where
[tex]dx[/tex] is the change in position
[tex]dt[/tex] is the change in time
If the position vs time graph of an object is a horizontal line, it shows that the object is at rest. At this point the object is not moving.
Answer:
If the position versus time graph of an object is a horizontal line, the object is at rest.
Reference:
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you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and direction of your average velocity
Answer:
Explanation:
We shall convert all the displacement in vector form .
i and j represents east and north respectively .
D₁ = 4 i
D₂ = 4 j
D₃ = - 5 cos 53.1 i + 5 sin 53.1 j
= -3i + 4 j
Total displacement = D₁ + D₂ + D₃
= 4i + 4 j - 3i + 4 j
= i + 8j
magnitude of displacement = √( 1² + 8² )
= 8.06 km
velocity = 8.06 / 5
= 1.61 km / h
Direction from x axis in anticlockwise direction .
Tanθ = 8 / 1 = 8
θ = 83° north of east .
If you walk at a constant speed of 4.5 m/s, how many km can you walk in half an hour? how do you solve this?
4. What is being carried from particle to particle in water wave?
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
tbh google
Your question has been heard loud and clear.
Energy propogates through different media in different ways.
Through a medium like water , energy is carried(propogated) from particle to particle in water waves.
Thank you.
A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth (Cable is going from the cage to ground). If the total mass of the balloon and the cage is 200 kg and the radius of the balloon is 5 m, how much tension (at least), the cable could be able to support in order to keep the balloon from breaking away. (Density of air is approximately 1.225 kg/m3 .)
Answer:
The tension is [tex]T = 4326.7 \ N[/tex]
Explanation:
From the question we are told that
The total mass is [tex]m = 200 \ kg[/tex]
The radius is [tex]r = 5 \ m[/tex]
The density of air is [tex]\rho_a = 1.225 \ kg/m^3[/tex]
Generally the upward force acting on the balloon is mathematically represented as
[tex]F_N = T + mg[/tex]
=> [tex](\rho_a * V * g ) = T + mg[/tex]
=> [tex]T = (\rho_a * V * g ) - mg[/tex]
Here V is the volume of the spherical helium filled balloon which is mathematically represented as
[tex]V = \frac{4}{3} * \pi r^3[/tex]
=> [tex]V = \frac{4}{3} * 3.142 *(5)^3[/tex]
=> [tex]V = 523.67\ m^3[/tex]
So
[tex]T = (1.225 * 523.67* 9.8 ) - 200 * 9.8[/tex]
[tex]T = 4326.7 \ N[/tex]
The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your school would be exactly twice as much if your school: a Had twice as much mass b Was twice as tall c Had twice as much volume d Covered twice as much area
Answer: a Had twice as much mass
Explanation:
The data that we have is:
"The force of gravity pulls down on your school with a total force of 400,000 newtons. "
First, remember that, by the second Newton's law that:
F = a*m
F = force
a = acceleration
m = mass
In the case of the gravitational force, the gravitational acceleration is a constant: a = 9.8m/s^2
Then, if we want to have twice as much force the only thing that we can change in the equation is the mass:
Then if the initial force is written as:
F = a*m
twice as much that force is written as:
2*F = a*x
x is a variable that represents the new mass.
We know that F = a*m
2*F = 2*a*m
2*a*m = a*x
2*m = x
Then, if we want to have twice as much force, we should have twice as much mass.
A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is his velocity at the end of this time
Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;
[tex]v = u + at[/tex]
where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes to a complete stop without bouncing back. The car has a mass of 1500 kg and has a speed of 30 m/s before the crash (this is about 65 mph). The driver has a mass of 60 kg.
1) What is the car's initial momentum? What is the person's initial momentum? What is the change in the momentum of the car? What is the change in the person's momentum?
2) What is the net impulse that acts on the car to bring it to a stop? What is the net impulse that acts on the person to bring them to a stop? Do these numbers depend on the details of the crash? Why or why not?
3) What object delivers the impulse that brings the car to a stop? What object delivers the impulse that brings the person to a stop? Think about what happens during such a crash, and discuss the role each of the factors below plays in the ultimate fate of the driver. Be sure to include what objects exert forces on what other objects, and use the model of impulse & momentum to estimate the strength of these forces.
a) "crumple zone" of the front of the car.
b) use or non-use of seat belt.
c) deployment or non-deployment of airbag.
Answer:
1) p₀ = 45000 N / s , p₀ '= 1800 , b) I = -45000 N s , I = 1800 Ns
Explanation:
Impulse equals the change in momentum
I = Δp
1) the initial moment of the car
p₀ = M v
p₀ = 1500 30
p₀ = 45000 N / s
the change at the moment is
Δp = 45000
because the end the car is stopped
moment of the person
P₀ ’= m v
p₀ '= 60 30
p₀ '= 1800
D₀ '= 1800
2) of the momentum change impulse ratio
car
I = Δp
I = -45000 N s
person
I = Δpo '
I = 1800 Ns
3) the object that give the momentum to stop the wall motoring
The person is stopped by the impulse given by the car
a) This area is the one that absorbs most of the vehicle impulse
be) If using a safety painter, the time during which the greater force will act, therefore the lessons decrease
c) The air bag helps reduction in the speed of the person relatively quickly.
An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x-direction when its x-coordinate is 2.94 cm. If its x-coordinate 1.85 s later is −5.00 cm, what is its acceleration?
Answer:
-19.9m/s²
Explanation:
Given that
vi = 14cm/s
xi = 2.94 cm
xf=-5.00 cm
t= 1.85s
So
Using xf-xi= vi t + 1/2at²
Which is = -5-2.94= 14*1.85+1/2a1.85²
= -7.94= 25.9+1.7a
-33.9= 1.7a
a= - 19.9m/s²
What is the Kinetic Energy of a pie with a mass of 5 kg travelling at a velocity of 9m/s towards your face ?
Answer:
K = 202.5 J
Explanation:
Given that,
Mass of pie is 5 kg
Velocity of pie is 9 m/s
We need to find the kinetic energy of a pie. The kinetic energy of an object is due to its motion. It can be given by the formula as follows :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 5\times (9)^2\\\\K=202.5\ J[/tex]
So, the kinetic energy of the pie is 202.5 J.
Using Newton's Second Law, can you explain why one of the major advancements in spaceflight was the development of strong ceramic (clay)
materials to build their rockets out of instead of heavy metals?
Answer:
high density can withstand high acceleration and applied forces
Heavy metals are toxic to humans,
the clay is quite abundant and in general it is not toxic
Explanation:
The selection of materials for the construction of rockets takes into account many aspects, the technical resistance to the demands of space travel, but also the abundance of the material. Heavy metals have two very serious problems. The first one, some of them are a little scarce in nature, but the most serious problem is that almost all of them are toxic to humans, for example: lead and mercury.
On the other hand, the clay is quite abundant and in general it is not toxic to living beings.
If we use Newton's second law
F = m a
let's use the concept of density
rho = m / V
m = rho V
let's substitute
F = rho V a
From this expression we see that a material with high density can withstand high acceleration and applied forces, such as those existing in spacecraft clearance and re-entry to Earth.
Unfortunately with this law there is no criterion to select a material unless its density is high, in addition to this criterion low toxicity criteria for human beings are used,
Determine the discharge of a river that is 10 m wide, has a semicircular cross-sectional area, and has a velocity of 3 m/s.
Answer:
117.8 m³/s
Explanation:
given that
the diameter of the river, d = 10 m wide
Velocity of flow of water, v, = 3 m/s
To get the discharge of a river, we have to find the area of the said river first.
Area of the river, a = area of semi circle = 1/2 * area of a circle
Area = 1/2 * πr²
Area = 1/2 * 3.142 * 5²
Area = 39.275 m²
Discharge of the river therefore is
Discharge, q = A * V, where
A is area of the river, and
V is the velocity of flow of the river.
Discharge, q = 39.275 * 3
Discharge, q = 117.825
What are the definitions of momentum, angular momentum, and kinetic energy for a single particle? What are the dimensions of these quantities?
Answer:
Explanation:
Momentum is defined as a physical quantity that takes into cognisance the product of mass of a body and its velocity. It's unit is kgm/s
Momentum = mv
Angular Momentum is defined as the quantity of rotation of a body that measures the product of its moment of inertia with respect to its angular velocity. Because it deals with rotation is what differs it from ordinary momentum. It's unit is kgm²/s
Angular Momentum = mvr
Kinetic Energy is defined as the energy a body posses as a result of it being in motion. The unit of Kinetic Energy is J
Kinetic Energy = ½mv²
The dimensions of these quantities are
Momentum => MLT^−1
Angular Momentum => ML²T^−1
Kinetic Energy => ML²T^-2
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What does F equal???.......ANSWER FAST A. ma B. m/a C. m+a D. m-a
Answer:
A. maExplanation:
Force = mass × acceleration
F = ma
(a)Suppose during 2020due to relief in Taxation Marginal Propensity to Consume (MPC) will increase from 0.65 to 0.75then calculate Tax multiplier (KT). (b)By using the Tax multiplier value calculated in part (a), find the change in national income if Tax amount is increased from 4500billion rupees to 5100billion rupees. (c)How Subsidies and Ehsas program amount will affect the Aggregate Demand (AD)?
Answer:
abcd
Explanation:
Answer:
where is the answer pls send me answer on my what.s up number
Explanation:
Physics help please..............
Answer:
3.68 m/s
Explanation:
Full answer in the attached picture
A 1200 kg racing car accelerates from rest at a constant rate and covers a distance of 450 m in 15 s. What is the car's acceleration
Answer:
2 m/s²
Explanation:
Given that
mass of the car, m = 1200 kg
Initial velocity of the car, u = 0 m/s
Distance covered, s = 450 m
Time taken, t = 15 s
Final velocity of the car, v = s/t = 30 m/s
there are four equations of motion. In this question, we will be attempting to solve it using the first equation of motion to solve this question
v = u + at, where
v = final velocity
u = initial velocity
a = acceleration
t = time taken
v = u + at
30 = 0 + 15a
30 = 15a
a = 30/15
a = 2 m/s
Which of the following surface environments is the most likely site for deposits in this photograph?
A. steep mountain front.
B. glacier.
C. sand dunes.
D. beach.
E. lake
Answer:
d. beach is the correct answer
Beach is the most likely site for deposits in this photograph. So, the correct option is (D).
What is Beach?A beach is defined as a narrow strip of land separating a body of water from inland areas usually made up of small grains of sand, rocks and minerals that have been eroded by frequent pounding by wind and waves. become The beach has both sandy and rocky features.
Beaches provide protection to residents living near the sea by acting as a buffer against high winds and powerful storms or waves of rough seas which also play an important role in the economy. This is the most likely site for deposits in this photograph,
Thus, Beach is the most likely site for deposits in this photograph. So, the correct option is (D).
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A continuous line of charge lies along the x axis, extending from x = +x0 to positive infinity. The line carries positive charge with a uniform linear charge density λ0. (a) What is the magnitude of the electric field at the origin? (Use the following as necessary: λ0, x0, and ke.) E = (b) What is the direction of the electric field at the origin? +x −x +y −y +z −z
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)