Answer:
Its traveling in the +x direction
Explanation:
The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.
The electromagnetic wave is travelling in the +x direction.
Electromagnetic waves are waves formed as a result of vibrations between an electric field and a magnetic field.
Given that:
Ey = E0 sin(kx - ωt)
Hence the electric field is moving in the +y direction.
Also, Bz = B0 sin(kx - ωt)
Hence the magnetic field is moving in the +z direction
Electric fields and magnetic fields (E and B) in an EM wave are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.
Therefore the direction of the wave is travelling in the +x direction.
Find out more at: https://brainly.com/question/25559554
A single slit 1.5 mm wide is illuminated by 420- nm light. Part A What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.5 m away
Answer:
The width is [tex]w_c = 0.00252 \ m[/tex]
Explanation:
From the question we are told that
The width of the single slit is [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 420 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 4.5 \ m[/tex]
Generally the width of the central maximum is
[tex]w_c = 2 * y[/tex]
where y is the width of the first maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{a}[/tex]
=> [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]
=> [tex]y = 0.00126 \ m[/tex]
So
[tex]w_c = 2 *0.00126[/tex]
[tex]w_c = 0.00252 \ m[/tex]
If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. This happens:_______
a. because of magnetic effects
b. because the ball tries to pull the rod's electrons over to it
c. because the rod polarizes the metal
d. because the rod and the ball have opposite charges
Answer:
c. because the rod polarizes the metal.
Explanation:
Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.
A sphere of radius R has charge Q. The electric field strength at distance r > R is Ei.
What is the ratio Ef /Ei of the final to initial electric field strengths if (a) Q is halved, (b) R is halved, and (c) r is halved (but is still > R)? Each part changes only one quantity; the other quantities have their initial values.
Answer:
A. Ef/ Ei = 1/2
B. EF/ Ei = 1
C Ef / Ei = 4
Explanation:
To solve this we apply Coulomb's law which States that
E = Kq / r^2
Where
q = charge r = straight line distance from q to the point in question and
K = Coulomb's constant
Then
Ei = K Q / r^2
So
A) If Q is halved then
Ef = K Q / (2 r^2)
Ef/Ei = 1/2
B) If R is halved, the value of the E-f
at a distance r remains unchanged. So
Ef/Ei = 1
C) if r is now r/2 then
Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2
Ef / Ei = 4
By what length will a slab of concrete that is originally 18 m long contract when the temperature drops from 24°C to -16°C? The coefficient of linear thermal expansion for this concrete is 1.0 × 10-5 C-1.
Explanation:
According to Thermal Expansion of solids:
[tex]dl = \alpha \times l \times dt[/tex]
[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]
[tex]dl = 7.2 \times {10}^{ - 3} [/tex]
How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet
Answer:
The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.
Fast please!!. Serious answers only.
Considering that Susan is 38 years old, what is the probability that she lives to the age of 85?
Answer:
The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.
Explanation:
Edmentum
Answer:
45,436/97,825= 0.4644.
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds. What is its rotation speed
Answer:
v = 6.28 m/s
Explanation:
It is given that,
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,
Number of revolution is half. It means angular velocity is 3.14 radians.
Let v is the angular speed. So,
[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]
So, the rotation speed is 6.28 m/s.
The angular velocity is the rotation speed, which is the angle of rotation
of the windmill per second, which is 2·π radians.
Response:
The rotation speed is 2·π rad/sHow can the rotational speed of the windmill be calculated?The given parameter are;
The angle of rotation the windmill rotates in 0.5 seconds = One-half a
rotation.
Required:
The rotational speed (angular velocity)
Solution:
The angle of one rotation = 2·π radians
Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians
[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]
Which gives;
[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]
The rotation speed is 2·π rad/sLearn more about rotational speed here:
https://brainly.com/question/6969329
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by:_____
a. a whole number of half-wavelengths.
b. a whole number of wavelengths.
c. an odd number of half-wavelengths.
Answer:
(B) a whole number of wavelengths.
Explanation:
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by a whole number of wavelengths.
When the resultant effect of the combination of two identical waves result in their annihilation or complete cancellation of the effect of each other, destructive interference takes place. Hence to have two wave sources producing waves that have the same frequency wavelength and amplitude and which are always in phase with each other or have a constant phase difference are said to be Coherent source
A projectile is launched with speed of 128 m/s, at an angle of 60° with the horizontal. After 2.0 s, what is the vertical component of the projectile's velocity?
After 2.0s, what is the speed of the projectile?
Answer:
a) 91 m/s
b) 111 m/s
Explanation:
v = u + at
v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s
v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s
A laser beam takes 15 ms to travel from a rocket to the reflective surface of a planet and back to the rocket. How far is the rocket from this planet's surface
Explanation:
It took 7.5 ms or [tex]7.5×10^{-3}\:\text{s}[/tex] for the laser beam to reach the planet's surface. Since the speed of light is [tex]3×10^8\:\text{m/s}[/tex], the distance is
[tex]d = vt = (3×10^8\:\text{m/s})(7.5×10^{-3}\:\text{s})[/tex]
[tex]\:\:\:\:=2.25×10^{6}\:\text{m}[/tex]
Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found
Answer:
only in or near star-forming clouds
Explanation:
When in the galactic disk, Red and orange stars are found evenly spread so here Blue stars are hot and therefore massive and therefore short-lived, that is means they never have time to venture far from the places, where they were born. so correct answer is blue stars are typically found only in or near star-forming clouds
During the first part of this lab, we want to determine how the object distance is related to what two quantities
Answer and Explanation:
The computation of the object distance related to two quantities is shown below:
It could find out by using the lens formula which is shown below:
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
where,
v = image distance
u = object distance
f = focal length
It could be found by applying the above formula i.e considering the image distance, object distance and the focal length
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic field at the center of the solenoid.a. 2.1E^-3T b. 1.0E^-3 T c. 1.7E^-4T d. 7.0E^-2 T
Answer:
The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Explanation:
The magnetic field B at the center of the solenoid is given by
B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.
So B = μ₀ni
= 4π × 10⁻⁷H/m × 1100 × 1.5 A
= 4π × 10⁻⁷H/m × 1650 A-turns/m
= 20734.5 × 10⁻⁷T
= 2.07345 × 10⁻³ T
≅ 2.1 × 10⁻³ T
So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answer:
Around the center of the mirror
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
If a soap bubble is 130 nmnm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated normally by white light
Answer:
The question is not complete, here is the other part.
Assume that n = 1.36.
Express your answer to three significant figures and include the appropriate units.
λ = 707.2nm
Explanation:
n = 1.36
t = 130
2 n t= (m+1/2) λ
To solve for λ.
λ = 2 n t ÷ m+1/2
λ = 2 × 1.36 × 130 ÷ m +1/2
λ =
If m= 0
λ= 353.6 ÷ 0+ 1/2
λ = 353.6 × 2
= 707.2nm
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?
Answer:
y = 12.82 m
Explanation:
We can solve this exercise using the energy work theorem
W = ΔEm
friction force work is
W = fr . s = fr s cos θ
the friction force opposes the movement, therefore the angle is 180º
W = - fr s
we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular
N -Wy = 0
N = mg cos θ
the friction force remains
fr = μ N
fr = μ mg cos θ
work gives
W = - μ mg s cos θ
initial energy
Em₀ = ½ m v²
the final energy is zero, because it stops
we substitute
- μ m g s cos θ = 0 - ½ m v²
s = ½ v² / (μ g cos θ)
let's calculate
s = ½ 20² / (0.55 9.8 cos 20)
s = 39.49 m
this is the distance it travels along the plane, to find the vertical distance let's use trigonometry
sin 20 = y / s
y = s sin 20
y = 37.49 sin 20
y = 12.82 m
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm
Answer:
40g
Explanation:
20g range > 1.0cm
Therefore,
40g range > 2.0cm
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a force F? =88i^?23j^ to the end of the wrench. What is the torque on the bolt?
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
?a wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).
Answer:
The speed of the rod is 2.169 m/s.
Explanation:
Given that,
Mass = 0.100 kg
Current = 15.0 A
Distance = 2 m
Length = 0.550 m
Kinetic friction = 0.120
(a). We need to calculate the magnetic field
Using relation of frictional force and magnetic force
[tex]F_{f}=F_{B}[/tex]
[tex]\mu mg=Bli[/tex]
[tex]B=\dfrac{\mu mg}{li}[/tex]
Where, l = length
i = current
m = mass
Put the value into the formula
[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]
[tex]B=0.01425\ T[/tex]
[tex]B=1.425\times10^{-2}\ T[/tex]
(b). If the friction between the rod and rail is reduced zero.
So, [tex]f_{f}=0[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F_{net}=f_{f}+F_{B}[/tex]
[tex]F_{net}=0+Bil[/tex]
[tex]ma=Bil[/tex]
[tex]a=\dfrac{Bil}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]
[tex]a=1.176\ m/s^2[/tex]
We need to calculate the speed of the rod
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value into the formula
[tex]v^2=0+2\times1.176\times2[/tex]
[tex]v^2=\sqrt{4.704}\ m/s[/tex]
[tex]v=2.169\ m/s[/tex]
Hence, The speed of the rod is 2.169 m/s.
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
A long, straight wire carries a 13.0-A current. An electron is fired parallel to this wire with a velocity of 250 km/s in the same direction as the current, 2.00 cm from the wire.
(a) Find the magnitude and direction of the electron’s initial acceleration.
(b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire?
(c) Is it necessary to include the effects of gravity? Justify your answer.
Answer:
A.5.7 x 10^-12m/s²
B. 32.5V/m
C. Pls. See attached file for explanations
Explanation:
An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is
[tex]E_{e} = E_{3} - E_{1}[/tex]
E[tex]_{e}[/tex] = initial kinetic energy of the electron
E[tex]_{1}[/tex] = -4 eV
E[tex]_{3}[/tex] = -1 eV
insert the values into the equation above
[tex]E_{e}[/tex] = -1 -(-4) eV
= -1 + 4 = 3 eV
Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.
Answer:
λ = 428.6 nm
Explanation:
Hello,
In this case, we must remember that the Young's double slit experiment is described by the expression :
d sin θ = m λ
For constructive interference , and:
d sin θ = (m + ½) λ
For destructive interference , whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference . Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation to compute the wavelength as shown below:
λ = 3x10⁻⁶ sin (30) / (3 +1/2)
λ = 4.286 10⁻⁷ m
Or in manometers:
λ = 428.6 nm
Best regards.
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction
Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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