Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.
Answers
Answer:
English only
Explanation:
When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:
1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ
The solutions:
A) Ey = 4623 N/C
B) Ey = 19.34 N/C
E = Ey = ∫ 2×dE× cosθ
Here cosθ = h/ d ⇒ cosθ = h/√h² + x² dE = K× dQ / d²
d² = h² + x²
k = 8.9 ×10⁹ Nm²C⁻² ; dQ = λ×dx λ = 150×10⁻⁹ C h = 0.08 m
Then by substitution
Ey = 2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²
reordering that equation:
Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³ (2)
To solve the integral we make use of a change of variables
x = h × tanα then x² = h² ×tan²α and dx = h× sec²α dα
plugging that values in equation (2)
Ey = 2×K×λ×h ∫ h× sec²α× dα / [√ ( h² + h²tan²α)]³
Ey = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³ 1 + tan²α = sec²α
Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα
Ey = 2×K×λ/h × ∫ ( 1 / secα dα
Ey = 2×K×λ/h × sinα now we αneed to come back to our original variables:
as x = h × tanα tanα = x/h then x is the opposite leg in a right triangle and h the adjacent one then the hypothenuse is √ (h² + x²) then sin α = x/ √ (h² + x²)
Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵
Ey = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 ) N/C
Ey = 4623 N/C
To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then
Ey = ∫ dE× cosθ
following the same procedure we will find:
Ey = ∫ [K×λ × dl/d²] × h/ d
The importan point here is that the radius of the circle is
2×π×r = 0.01 ( the length of the wire) ⇒ r = 0.16×10⁻² m
And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution
Ey = [K×λ ×h× / ( √ r² + h²)³ ] × 10⁻² N/C
Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)
Ey = 0.8 × 10² / 6
Ey = 19.34 N/C
Related Questions
The brakes in car A and car B stop working and they need to stop. Car A and car B have the same mass. Car A is traveling at 30 miles per hour, smashes into a big wall, and is brought to rest in a collision that lasts for a fraction of a second. Car B was also traveling at 30 mph but it drives out into a field of grasses and sand and finally comes to rest after a bumpy 20 second ride. Which car gets the larger impulse?
Answers
Answer:
Both have equal impulse.
Explanation:
Let the mass of cars be m.
Then the Force acting on each of them for taking them to state of rest:
(Using Newton's second law of motion)
[tex]F_A=\frac{m\times (0-30)}{\Delta t_A}[/tex]
[tex]F_A=-30\frac{m}{\Delta t_A}[/tex] ...................................(1) (negative sign is associated with direction, here we are concerned about the magnitude only)
[tex]F_B=\frac{m\times (0-30)}{\Delta t_B}[/tex]
[tex]F_B=-30\frac{m}{\Delta t_B}[/tex] ...................................(2)
[tex]\because \Delta t_A<\Delta t_B[/tex]
[tex]\therefore F_A>F_B[/tex]
We know that impulse is given as:
[tex]J=F\times \Delta t[/tex] ........................................(3)
So, from eq. (1), (2) & (3)
[tex]J_A=F_A\times \Delta t_A[/tex]
[tex]J_A=-30m[/tex]
&
[tex]J_B=F_B\times\Delta t_A[/tex]
[tex]J_B=-30m[/tex]
Hence both have equal impulse.
If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply
Answers
Answer:
[tex]T=8.33A[/tex]
Explanation:
From the question we are told that:
Number of battery [tex]n=10[/tex]
Voltage source[tex]E=12V[/tex]
Lamp Power[tex]P=10W[/tex]
Generally the equation for Resistance is mathematically given by
[tex]R=\frac{V^2}{P}[/tex]
[tex]R=\frac{12^2}{10}[/tex]
[tex]R=14.4ohms[/tex]
Therefore
[tex]R_{eq}=\frac{14.4}{10}[/tex]
[tex]R_{eq}=1.44[/tex]
Generally the equation for Current is mathematically given by
[tex]T=\ffrac{V}{Req}[/tex]
[tex]T=\frac{12}{1.44}[/tex]
[tex]T=8.33A[/tex]
A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squirrel?
Answers
Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
[tex]v_{0y}=4.0sin(50.0)[/tex] which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]3.1(.32)+\frac{1}{2}(-9.8)(.32)^2[/tex] and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
Name the fundamental units involved in the derived unit joule.
Answers
Explanation:
[tex]we \: know \: that \: joule \: is \: the \: unit \: of \\ energy \: we \: have \: energy = \frac{1}{2} m {v}^{2} \\ unit \: of \: this \: is \\ joule = kg {( \frac{m}{ s }) }^{2} \\ thank \: you[/tex]
1.8kg 42J 9.8 how high is the shelf
Answers
Answer:
2.38m
Explanation:
Use potential energy
PE= mgh
42= 1.8*9.8*h
solve for h
to get h= 2.38 m
A scientist that applies the laws of science to the needs of communities is called _____.
the experimental scientist
the engineer
the teacher
the technician
Answers
Answer:
The experimental scientist
An acceptable radiograph was taken using 40 mAs at 80 kVp at a distance of 60 inches. A second radiograph is requested at 40 inches. What mAs should be used to produce this radiograph with a 40-inch distance
Answers
Answer:
17.8 mAs
Explanation:
The exposure maintenance formula shows that as SID increases, intensity decreases, causing a decrease in film exposure and density. The mAs is directly proportional to the square of the distance. That is as mAs increases, distance increases, and vice versa, in order to maintain image receptor exposure. It is given by:
[tex]\frac{mAs_1}{mAs_2}=\frac{D_1^2}{D_2^2}[/tex]
Given that mAs₁ = original mAs = 40 mAs, D₁ original distance = 60 in, D₂ = new distance = 40 in, mAs₂ = new mAs
[tex]\frac{mAs_1}{mAs_2}=\frac{D_1^2}{D_2^2}\\\\mAs_2=\frac{mAs_1*D_2^2}{D_1^2}=\frac{40*40^2}{60^2}=17.8\\\\mAs_2=17.8\ mAs[/tex]
a 2kg object is dropped from height of 10m. ignoring air resistance calculate:
1. mechanical energy of the object
2. kinetic energy of the object when it is 3m above the ground
Answers
Answer:
ME= 196.2 J
KE= 136.2
Explanation:
potential energy=mgh 2*9.81*10
Our ME is quivalent to PE as that is the toal amount of energy in the system
Kinetic energy= 1/2 m[tex]v^{2}[/tex]
to solve for kinetic enrgy we need to use a kinaetmtic equation that help us find velocity
vf= vi+at
but we need to find time first
d=vi+1/2(accelretaion)[tex]t^{2}[/tex]
7=0+1/2(9.81)[tex]t^{2}[/tex]
t= 1.19 s
vf= 0+ 9.81*1.19
vf= 11.67 m/s
Now
1/2 m[tex]v^{2}[/tex]
1/2*2*[tex]11.67^{2}[/tex]
= 136. 2
or we could just (PE/10)*7
so (196.2/10)*7
What is the current flowing through the circuit shown? (V= 110 V, R, = 200, R2 = 300, R3 = 10 0) (Ohm's law: V = IR)
A. 1.8 A
B. 20 A
C. 0.05 A
D. 0.55 A
Answers
Answer:
A
Explanation:
The first thing you have to do is go back and list the resistances correctly. R1 = 20 R2 = 30 and R3 = 10.
Leave the units off if you can't make an omega.
The resistance of a series circuit (that's what this is) is r1 + r2 + r3 = 10 + 20 + 30 = 60 ohms
Now use ohms law.
R = 60 ohms
V = 110 volts.
I = V / R
I = 110/60
I = 1.833 to the nearest 1/10 = 1.8
If mass is the amount of MATTER present in the
object, will it be different on Earth and the moon?
No
Yes
Answers
Answer:
Yes it would be different on Earth and the moon
Answer:
yes
Explanation:
(c) The metal block had a mass of 1.50 kg
The specific heat capacity of the metal was 900 J/kg °C
Calculate the change in thermal energy of the metal during 240 seconds.
Use the Physics Equations Sheet.
Give your answer in kilojoules.
Answers
Answer:56
Explanation: I did the test
prove that vector addition obeys commutative and associative laws
Answers
the order of vector addition doesn't affect the resultant vector and grouping or order of pair doesn't effect the sum.
Mean diameter of cylinder measured by A is 2.50 cm with a possible error of 0.25 cm whereas mean height of cylinder measured by B is 10.0 cm with a possible error of 0.5 cm. Whose measurement is more accurate?
Answers
Answer:
The measurement of first cylinder is more accurate.(A)
Explanation:
The least possible error higher will be accuracy.
The weight of a body is 600 N. What is the mass of the body on the surface of the earth?
Answers
Explanation:
soln,
weight=600N
mass=?
gravity=9.8 m/s²
now,
mass=weight/gravitymass=600/9.8mass=61.22kghope it helps.
stay safe healthy and happy.Answer:
m = 61.22 kg
Explanation:
F = 600 N
g = 9.8 m/s²
m = ?
F = mg
600 = m(9.8)
---> m = 61.22 kg
What is the correct equation for the y axis of object B?
Answers
Answer:
[tex]N_B-W_B = 0[/tex]
Explanation:
There are two forces acting on object B. If we consider the law of equilibrium, then the two forces must be equal in magnitude, for the object to remain in equilibrium position:
[tex]N_B = W_B\\\\N_B-W_B = 0[/tex]
Therefore, the correct answer of the given question, from the given choices, is:
[tex]N_B-W_B = 0[/tex]
A fast moving car is moving on a straight road with an acceleration of 2m/s2. How much time will it require to change the velocity from 20m/s to 40m/s.
Please quick
Answers
Answer:
10seconds
Explanation:
use the formula a=v final_v inital/time
Richard rides a bike on a road sloping downward and inclined at an angle of 30° with the horizontal. If he moves at a constant speed of 10 meters/second, calculate his vertical downward velocity.
Answers
Answer:
[tex]v_y = 5\ m/s[/tex]
Explanation:
The vertical downward component of the speed of the bike can be simply found out by using the following geometric formula:
[tex]v_y = vSin\theta[/tex]
where,
[tex]v_y[/tex] = vertical downward component of speed = ?
v = speed = 10 m/s
θ = angle of slope = 30°
Therefore,
[tex]v_y = (10\ m/s)Sin\ 30^o\\v_y = 5\ m/s[/tex]
(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface is 0.412 m/s2
Answers
Answer:
[tex]V.E=498.02m/s^2[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=301Km[/tex]
Gravitational acceleration [tex]g=0.412 m/s^2[/tex]
Generally the equation for Escape velocity is mathematically given by
[tex]V.E^2=2gr[/tex]
[tex]V.E^2=2*0.412m/s^2*301000[/tex]
[tex]V.E^2=248024[/tex]
[tex]V.E=\sqrt{248024}[/tex]
[tex]V.E=498.02m/s^2[/tex]
URGENT A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes. a. What was his average speed? b. What was his displacement? PLEASE SHOW ALL WORK
Answers
Answer:
Explanation:
As far as the displacement goes, we have 2 displacement vectors. If we didn't have the angles to deal with, this would be a much simpler process, but then that wouldn't be any fun at all, would it? I'll deal with the average speed first, then the displacement, which is a vector addition problem.
The average speed is found by adding together the distances the student traveled and then dividing this sum by the total time he spent traveling. If we are told that the student runs at 4.5 m/s for 3.0 minutes, we can use this to find out the distance he ran during that time interval. However, the units are not the same. We will find the distance the student traveled by convering the time to seconds.
3.0 minutes = 180 seconds, and
4.1 minutes = 246 seconds.
That means that the distance he ran in 180 seconds is found by multiplying this time be the speed at which he ran:
4.5 m/s(180 s) = 810 m and
3.5 m/s(246 s) = 860 m (rounded to follow the rules of sig dig).
This makes the speed equation look like this:
[tex]s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}[/tex] That's the average speed, which is NOT at all the same as the displacement. Displacement is where he ended up in reference to where he started. The angles play a huge part in this math (that is very involved, to say the least). We begin by restating the displacement of each "leg" of this journey.
The first leg took him 810 m at 207 degrees and
the second leg took him 860 m at 325 degrees
To find the x and y components of these 2 legs, or parts, we have to use the cos and sin formulas. We will call the first leg A and the second leg B. First the x components of both A and B:
[tex]A_x=810cos207[/tex] and
[tex]A_x=-720[/tex]
[tex]B_x=860cos325[/tex] and
[tex]B_x=704[/tex] and we add these to get the x-component of the resultant vector, C:
-720
+ 704
-10 (rounded, as needed, to the tens place).
Now for the y-components of the resultant vector:
[tex]A_y=810sin207[/tex] and
[tex]A_y=-370[/tex]
[tex]B_y=860sin325[/tex] and
[tex]B_y=-490[/tex] and we add these to get the y-component of the resultant vector, C:
-370
+ -490
-860
Since the x component is negative and so is the y, we are in QIII, so when we finally find our angle, we will have to add 180 to it.
For the magnitude of the displacement vector, in m:
[tex]C_{mag}=\sqrt{(-10)^2+(-860)^2}[/tex] which gives us
[tex]C_{mag}=860m[/tex]
Now, because displacement is vector, we also need the angle. We find that is the formula
[tex]\theta=tan^{-1}(\frac{C_y}{C_x})[/tex] and filling in:
[tex]\theta=tan^{-1}(\frac{-860}{-10})=90[/tex] (rounded correctly), and then we add 180 to give us a final direction of 270 degrees.
So the final displacement of the student is 860 m at 270 degrees
Clothes are flung to the sides of a round tub during the spin-cycle of a washing machine.If the tub's period of rotation is 0.2s and the tub's radius is 35cm,answer the related qestions below: (a)The frequency of roation (b)The circumference of rotation (c)The spinning of speed (d)The centripetal acceleration (e)The centripetal force of rotation if total mass rotated is 30kg (f)The power of washing machine
Answers
Answer:
Explanation:
First I need to tell you that I used .20 s for the period of rotation instead of just .2, and I used 30.0 kg for the mass instead of just 30. The reason being that both those numbers as stated in the problem only have 1 significant digit and that's not generally enough to get the accuracy you're looking for. Adding a 0 to the ends of each of those numbers doesn't change the value of the numbers, only the number of sig fig's in each. Beginning with a:
a. [tex]f=\frac{1}{T}[/tex] so [tex]f=\frac{1}{.2}[/tex] and f = 5.0 Hz
b. The circumference is the distance around the outside of the washer's drum. We need to find that, but before we do, I'm going to state the radius in meters instead of cm. 35 cm = .35 m. Therefore,
C = 2(3.1415)(.35) so
C = d = 2.2 m
c. The speed of the washer is found in d = rt, where r is the rate and our velocity and d is the distance around the outside of the drum (circumference). Therefore,
2.2 = v(.20) so
v = 11 m/s
d. The centripetal acceleration has an equation
[tex]a_c=\frac{v^2}{r}[/tex] so
[tex]a_c=\frac{(11)^2}{.35}[/tex] and
[tex]a_c=\frac{121}{.35}[/tex] so
[tex]a_c=350\frac{m}{s^2}[/tex]
e. The centripetal force has an equation
[tex]F_c=\frac{mv^2}{r}[/tex] and
[tex]F_c=\frac{(30.0)(11)^2}{.35}[/tex] and
[tex]F_c=[/tex] 1.0 × 10⁴ N
f. The equation for Power is
[tex]P=\frac{W}{t}[/tex] where W is work and W = FΔx (force times displacement). Therefore,
[tex]P=\frac{(1.0*10^4)(2.2)}{.20}[/tex] so
P = 1.1 × 10⁵ Watts
In 'coin on card' experiment a smooth card is used.
Answers
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass
What is the size of the force acting on a copper wire with a magnetic flux density of 3.6 x 10-2 T acting at
right angles to the wire of length 24 m and current of C:048 A? Give your answer to an appropriate
number of significant figures.
Answers
Answer:
F = 0.414 N
Explanation:
Given that,
Magnetic flux density,[tex]B=3.6\times 10^{-2}\ T[/tex]
The length of the wire, l = 24 m
Current, I = 0.48 A
We need to find the force acting on the wire. The formula for the force is given by:
[tex]F=ILB[/tex]
Put all the values,
[tex]F=0.48\times 24\times 3.6\times 10^{-2}\\\\F=0.414\ N[/tex]
So, the force acting on the copper wire is equal to 0.414 N.
The magnetic force of the copper wire is 41.472 N.
Magnetic force of the copper wire
The magnetic force of the copper wire is calculated by applying the following equation.
F = BIL x sinθ
Where;
θ is the inclination of the magnetic fieldI is the currentL is the length of the wireB is the magnetic field strength = flux densityF = (3.6 x 10⁻²) x (48) x 24 x sin(90)
F = 41.472 N
Thus, the magnetic force of the copper wire is 41.472 N.
Learn more about magnetic force here: https://brainly.com/question/13277365
Which two factors affect the size of the gravitational field?
Answers
Answer:
Explanation:
mass and distance
What is an electric fuse? What is the working principle of electric fuse?
Answers
A system consists of multiple objects connected by ropes. How many equations need to be written to solve this problem?
A) two for each object
B) one or two for each object
C) one for each object
D) two for the system
Answers
Answer:
the correct answer is B
Explanation:
To solve the system they must have the same amount of unknowns as equations,
a) If the system does not have friction, we must write the x-axis equation for each body, therefore we need to write N equations
b) if the system has friction, two equations are needed for each particle
therefore the correct answer is B
How does the construction of dams positively affect natural resources?
by providing water for irrigation and restoring trees to areas where forests once existed
by creating reservoirs, preventing flooding, and renewing destroyed ecosystems
by preventing flooding, creating reservoirs, and providing water for irrigation
by renewing destroyed ecosystems and restoring trees to areas where forest once existed
Answers
Other person is incorrect, I did the test on edg.
Answer:
A. by providing water for irrigation and restoring trees to areas where forests once existed
The construction of dams positively affect natural resources by providing water for irrigation.
What is natural resources?Natural resources serves as materials thst is found in nature which are used by humans. These natural substances are soil, water.
Therefore, building of dams can help irrigation by using water which is a natural resources.
Learn more about natural resources at;
https://brainly.com/question/24623157
Assertion: When I P+ Q I = I P- QI, then P must be perpendicular to Q. Reason: The relation will hold even when Q is a null vector.
Answers
Answer:
The assertion is true and reason is false.
Explanation:
Assertion: I P+ Q I = I P- QI, then P must be perpendicular to Q.
Reason : The relation will hold even when Q is a null vector.
Now
[tex]\left | P + Q \right |=\left | P - Q \right |\\\\P^2 + Q^2 + 2 P Q cos \theta =P^2 + Q^2 - 2 P Q cos \theta\\\\4 P Q cos \theta = 0 \\\\cos \theta = 0 \\\\\theta = 90 degree[/tex]
So, P and Q are perpendicular to each other.
So, the assertion is true.
Reason is false.
what happens to gravitational force when distance is quarter from the original
please answer it faster help me
Answers
Answer:
The force of gravitational attraction between them also decreas
the electric charge remains at rest in a. static electricity, b. current electricity, c. none of these
Answers
Answer:
Option (a) is correct.
Explanation:
The electric charge is the inherent property of matter which appears due to the deficiency of charge or the excess number of electrons.
There are two types of charges, positive and negative.
When an object has deficiency of charge it becomes positive charge and when it gains some electrons, it becomes negative charge.
When the charge is in motion, it is called electric current and when the charge is at rest, it is called static current.
A charge at rest contributes the electric field only while the charge is in motion contributes the electric and magnetic field both.
The Earth orbits around the sun because the gravitational force that the sun
exerts on the Earth:
O A. causes Earth's acceleration toward the sun.
O B. is very small because the sun is so far from the Earth.
O c. is smaller than the force the Earth exerts on the sun.
O D. pushes the Earth away from the sun.
Answers
The Earth orbits around the sun because the gravitational force that the sun
exerts on the Earth:
O A. causes Earth's acceleration toward the sun.
O B. is very small because the sun is so far from the Earth.
O c. is smaller than the force the Earth exerts on the sun.
O D. pushes the Earth away from the sun.
Answer -:O A. causes Earth's acceleration toward the sun.
I hope this helps, have a nice time ahead!
