Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun.
Estimate gravity on the surface on this star.
I know the solution of this question it is as the picture shows but I only need to add *10^3 to the lower part of the division lower part to get the correct answer. But I don't know why I should add it can anyone explain?
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
[tex]g=4.38*10^{7} m/s^2[/tex]
Explanation:
To solve this, we need to know the mass of the sun, and the radius of the moon
[tex]M_{s} = 1.989*10^{30}kg \\R_{m} = 1737400m[/tex]
Now we can plug our values into our equation:
[tex]g=G*\frac{M_{E} }{r^{2} }[/tex]
This gives us:
[tex]g=6.67*10^{-11}*\frac{1.989*10^{30}}{1737400^{2}}[/tex]
This equals:
[tex]g=4.38*10^{7} m/s^2[/tex]
describe why people are better off not consuming an additional good or service if the marginal cost is greater than the marginal benefit.
As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then convected to the surrounding air. The steady-state temperature of the wire can be computed as
Answer:
Hello your question is incomplete attached below is the complete question
answer : ri ( thickness of wire ) = 14.167 m
Explanation:
attached below is a detailed solution
using the given data to determine the thickness of wire
As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy
Answer:
isnt heat transfer
Explanation:
sorry if im wrong
How does speed and mass effect kinetic energy ?
Answer:
in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.
Explanation:
Answer: The more mass of an object has, the more Kinetic energy it has.
Explanation:
Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.
A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B with silk. Nylon is used to rub and charge a third piece of amber. Piece A and B are both repelled by the third piece of amber. This means:____.
Answer:
ieces A and B must also have the same type of charges
Explanation:
In electrostatics, charges of the same sign repel and charges of different signs attract.
If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.
Also part A and C repel each other, therefore they have the same type of charge.
If we use the transitive property of mathematics, pieces A and B must also have the same type of charges
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?
Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec
differentiate between computer and computer system
A computer is a programmable device that can automatically perform a sequence of calculations or other operations on data once programmed for the task. It can store, retrieve, and process data according to internal instructions. A computer may be either digital, analog, or hybrid, although most in operation today are digital. Digital computers express variables as numbers, usually in the binary system. They are used for general purposes, whereas analog computers are built for specific tasks, typically scientific or technical. The term "computer" is usually synonymous with a digital computers, and computers for business are exclusively digital.
Answer:
The core, computing part of a computer is its central processing unit (CPU), or processor. ... A computer system, therefore, is a computer combined with peripheral equipment and software so that it can perform desired functions.
Explanation:
Hope the answer was helpful
Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is different and your example
Please help due today!!
Answer:
The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.
Examples of Gravitational Energy
A raised weight.
Water that is behind a dam.
A car that is parked at the top of a hill.
A yoyo before it is released.
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
Einstein's equivalence principle says that __________. Einstein's equivalence principle says that __________. everyone measures the speed of light to be equivalent someone traveling at 0.9c will age at the same rate as someone at 0.99c all people see themselves at an equivalent distance to the center of the universe the effects of gravity are exactly equivalent to the effects of acceleration
Answer:
Einstein's equivalence principle says that __________.
the effects of gravity are exactly equivalent to the effects of acceleration
Explanation:
The equivalence principle is one of the fundamental laws of physics, as enunciated by Einstein. It categorically states that the gravitational and inertial forces are of a similar nature. In physics, a gravitational acceleration is the acceleration of an object in a free fall within a space. The importance of Einstein's Equivalence Principle is explained by his theory of general relativity. This theory states that mass is the same, whether inertial or gravitational.
According to the Einstein's equivalence principle, the effects of gravity are exactly equivalent to the effects of acceleration.
Einstein's equivalence principle says that the effects of gravity are exactly equivalent to the effects of acceleration.
What is Einstein's equivalence principle?Einstein's equivalence principle states that the the force due to gravity and the force of inertia are similar in the nature and there is no need to distinct them.
The inertia force is opposite in direction to accelerating force of a body. Thus the Einstein's equivalence principle can also be stated as "the effects of gravity are exactly equivalent to the effects of acceleration." Form the given option the correct option which can be filled in the blank is option 2 which states that the effects of gravity are exactly equivalent to the effects of acceleration.Thus Einstein's equivalence principle says that the effects of gravity are exactly equivalent to the effects of acceleration.
Learn more about the Einstein's equivalence principle here
https://brainly.com/question/20533998
A 9800 N car traveling at 22 m/s strikes a concrete bridge support and comes to a complete halt in 0.5 sec. (This only counts as one problem.) Determine the magnitude of the force acting on the car. Suppose a barrier, which contained material that is gradually crushed during impact so that the stopping time was increased to 3 s, had surrounded the bridge support. What would be the magnitude of the new force
Answer:
43955.12N and 7325.85N
Explanation:
Step one:
Given data
Weight of car= 9800N
therefore the mass of the car is
w=mg
m= w/g
m= 9800/9.81
m= 998.98 kg
Speed= 22m/s
duration of impact= 0.5 seconds
we know that
impulse, Ft= mv
F= mv/t
F= 998.98*22/0.5
F=21977.56/0.5
F=43955.12N
Hence the force is 43955.12N
if the time is 3 seconds
then the force will be
F= mv/t
F= 998.98*22/3
F=21977.56/3
F=7325.85N
At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himself. Mason pulled the bat with
force of 15 newtons (N) while Alfredo pulled with a force of 20 newtons (N). Why did Alfredo end up with the bat?
A because the force was 5 N in Mason's direction
B. O because the net force was 5 N in Alfredo's direction
c. O because the net force was 15 N in Mason's direction
D.O because the net force was 20 N in Alfredo's direction
Answer:
Option B. O because the net force was 5 N in Alfredo's direction
Explanation:
To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:
Force of pull by Mason (Fₘ) = 15 N
Force of pull by Alfredo (Fₐ) = 20 N
Net force (Fₙ) =?
Fₙ = 20 – 15
Fₙ = 5 N in Alfredo's direction
From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V
Answer:
[tex]E=6.8Kv/m[/tex]
Explanation:
From the question we are told that
Distance b/w plate [tex]d=10cm=>0.1m[/tex]
P_1 Potential at 7.35 [tex]V=533v[/tex]
Generally the equation for electric field at a distance is mathematically given as
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{533}{7.85*10^-^2}[/tex]
[tex]E=6789.808917[/tex]
[tex]E=6.8*10^3[/tex]
[tex]E=6.8Kv/m[/tex]
An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12 cm from it, there is a converging lens of the same focal length.
A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?
Answer:
A) q = -8.488 cm , B) m = 0.29
Explanation:
A) For this exercise in geometric optics, we will use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where p and q are the distance to the object and image, respectively and f is the focal length
in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm
[tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]
we calculate
[tex]\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}[/tex]
[tex]\frac{1}{q}[/tex] = - 0.1178
q = -8.488 cm
the negative sign indicates that the image is virtual
B) the magnification is given
[tex]m = \frac{h'}{h} = - \frac{q}{p}[/tex]
we substitute
m = [tex]- \frac{-8.488}{29}[/tex]
m = 0.29
the positive sign indicates that the image is right
1.18. Which of the following is/are supplementary unit(s)? (1) Kelvin
(II) Newton (III) Second (IV) Radian
A. I and III only
C. I and II only
B. IV only
D. I, II and IV only
Answer:
B. IV only
Explanation:
Ellie is shooting a free throw. At the instant the ball leaves her hand the ball is traveling at a velocity of 9 m/sec, at an angle above the horizontal of 29.5 degrees. If you are standing beneath the basketball rim, videotaping, you can only see the vertical motion of the ball, since its horizontal motion is straight towards you. What is the magnitude of the ball's initial vertical velocity (if you measured it using your video data, in m/sec)
Answer:
Vy = 4.43 m/s
Explanation:
given data
velocity = 9 m/sec
angle horizontal = 29.5 degrees
solution
we get here ball initial vertical velocity will be here
Vy = v sinθ ...........................1
put here value and we get
Vy = 9 × sin(29.5 )
Vy = 4.43 m/s
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 865 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed. You find that the forces are attractive and the magnitude of each force is:______
Answer:
force F = 1.66 × [tex]10^{-13}[/tex] N
Explanation:
given data
proton and an electron = 865 nm
solution
we get here force that is express as
force F = k q1 q2 ÷ r² ......................1
put here value and we get
force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]
force F = 1.66 × [tex]10^{-13}[/tex] N
In the Bohr model of the atom, the electrons are considered to be particles moving in specific and discrete orbits.
a. True
b. False
Answer:
True
Explanation:
Bohr proposed an atomic model in which;
the electrons found in an atom can only occupy orbits for which the angular momentum is quantized, which results in discrete values of energy level. the electrons in such stationary state or ground state emits no light, but if an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states.18.) Why is it that Prokaryotic cells can divide a lot quicker than Eukaryotic cells?
Answer:
Binary Fission. The cell division process of prokaryotes, called binary fission, is a less complicated and much quicker process than cell division in eukaryotes. Because of the speed of bacterial cell division, populations of bacteria can grow very rapidly.
Work put into a machine can never be reclaimed.
True
False
Please answer?!
I am not sure but I think it is false
In the Faraday pail experiment a metal ball is lowered into a brass pail. In a variation of this experiment, suppose we charge the metal ball with Q and the brass pail with charge 2Q. The ball is slowly lowered into the pail. At no time does the sphere touch the pail. While the ball is inside, the charge on the outside of the pail is _____. The outer part of the pail is then touched/grounded. Then the ball is removed. After the ball is removed, the charge on the outer surface of the pail is ___. Briefly support your answers with reasoning from Gauss's Law.
Answer:
a)
*The charges of which and the cube are of the same sign..
Q_exterior = 3 Q
* the charge of the sphere has a different sign than the charge of the cube,
Q_exterior = Q
b) Q = 0
Explanation:
To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.
Let's analyze each situation
a) Suppose that the charges of which and the cube are of the same sign.
When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of
Q_exterior = 3 Q
Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,
Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall
Q_exterior = + Q
b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge
Q = 0
Suppose that material scientists are testing a new alloy that has remarkably large coefficients of thermal expansion. A sample of the alloy is formed into a thin, uniform, solid disk with diameter 2.00 cm when at room temperature. If the temperature of the disk is increased by an amount ΔT the diameter of the disk increases by an amount ΔD.
a. True
b. False
Answer:
True
Explanation:
When the temperature of solid is increased its size increases. It is due to the increase in internal energy of the solid and then there is expansion in the solid. The temperature gives heat to the solid object and it expands in size. The expansion cannot be linear it can be increased in size by the volume.
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
Which statement about technology is true?
Technology improves a scientist’s ability to make observations.
Technology does not have any effect on a scientist’s work.
Technology is necessary for scientists to perform experiments.
Technology only helps scientists analyze data and write reports.
Answer:
Technology improves a scientist’s ability to make observation
hope this helps
:D
Answer:
Technology improves a scientist’s ability to make observations.
Explanation:
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Answer:
average speed is 1 m/s and average velocity is 0 m/s.
Explanation:
Given that :
Length of round trip = 50 m
Time taken = 100 seconds
The average speed :
Total distance / total time taken
Length of complete round trip :
(50 + 50) m, total. Distance = 100 m
100 / 100 = 1m/s
The average velocity :
Total Displacement / total time taken
Total Displacement of round trip = end point - start point = 0
0 / 100 = 0
Average speed is 1 m/s and average velocity is 0 m/s.
The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.
The speed is obtained from;
Speed = Distance/time = 2(50 m)/100 s = 1 m/s.
The velocity is 0 m/s since it is complete round-trip lap.
Learn more about speed: https://brainly.com/question/7359669
The height above the ground of a stone thrown upwards is given by s(t), where t is measured in seconds. After 1secondnothing,the height of the stone is 47feet above the ground, and after 1.5seconds, the height of the stone is 54feet above the ground. Evaluate s(1)and s(1.5),and then find the average velocity of the stone over the time interval [1,1.5].
Answer:
v = 14 ft/s
Explanation:
By definition, the average velocity, is just the rate of change of the position, with respect to time, which can be written as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_{f}-x_{o}}{t_{f} - t_{o}} (1)[/tex]
Defining the vertical position as the y-coordinate, with the origin at ground level, and the upward direction as positive, we can write (1) as follows:[tex]v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} (2)[/tex]
where yf = 54 ft, y₀ = 47 ft, tif = 1.5 s, t₀ = 1s.Replacing in (2) we get:[tex]v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} = \frac{7ft}{0.5s} = 14 ft/s (3)[/tex]
Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.
Complete Question
The question image is in the first uploaded image
Answer:
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
[tex]E=\frac{kQ}{d^2}[/tex]
where
[tex]k=\frac{1}{4\pi e_0}[/tex]
[tex]d=x^2-a^2[/tex]
Therefore
[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]
[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]
Therefore equation for magnitude of the electric field at the point P is
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]