Answer:
Poynting vector oscillate at a frequency of 2omega
Explanation:
This is because The poynting vector is proportional to the cross product of electric and magnetic field vectors. So Because both fields oscillate sinusoidally with frequency w, trigonometric identities show that their product is a sinusoidal function of frequency of 2w.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? =
Answer:
x = -3 cm
Explanation:
The electrical potential is the sum of the potentials of each charge
V = k ∑ [tex]q_{i} / r_{i}[/tex]
let's apply this to our case where the potential is V = 0 for x = 0
0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))
in our case
q₁ = + 2.0 10⁻⁶ C
q₂ = - 6.0 10⁻⁶ C
q₃ = + 3.0 10⁻⁶ C
x₁ = -1.0 cm = 1.0 10⁻² m
x₂ = +2.0 cm = 2.0 10⁻² m
we substitute in the equation
0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)
3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴
3 10⁻⁶ / x = -1 10⁻⁴
x = - 3 10⁻² m
x = -3 cm
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
A 0.10 kg point mass moves in a circular path with a radius of 0.36 m with a net force of 10.0 N toward the center of the circle. Select all of the following that are true statements.
a. The velocity of the object is 6 m/s toward the center of the circle.
b. The speed of the object is 6 m/s and decreasing.
c. The speed of the object is 6 m/s and increasing.
d. The velocity of the object is a constant 6 m/s.
e. The speed of the object is a constant 6 m/s.
Answer:
e. The speed of the object is a constant 6 m/s
Explanation:
Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .
Centripetal force = m v² / r , r is radius of circular path .
Putting the given values
.10 x v² / .36 = 10
v = 6 m /s
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 480 turns, its secondary 8 turns, and the input voltage is 118 V
Answer:
1.97 V
Explanation:
Applying
N1/N2 = V1/V2................... Equation 1
Where N1 = primary turns, N2 = Secondary turns, V1 = primary/input voltage, V2 = secondary/output voltage.
make V2 the subject of the equation
V2 = V1(N2/N1)............. Equation 2
Given: V1 = 118 V, N1 = 480 turns, V2 = 8 turns.
Substitute into equation 2
V2 = 118(8/480)
V2 = 1.97 V.
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
A person takes a trip, driving with a constant speed of 98.5 km/h, except for a 20.0-min rest stop. The person's average speed is 68.8 km/h. (a) How much time is spent on the trip? h (b) How far does the person travel? km
Answer:
Total time taken(T) = 1.1 hour
Distance = 75.68 km
Explanation:
Given:
Average speed = 68.8 km/h
Constant speed = 98.5 km/h
Rest time = 20 min = 20 / 60 = 0.3333 hour
Find:
Total time taken(T)
Total distance (D)
Computation:
Distance = speed × time
D = 68.8 × t.........Eq1
and
D = 98.5 × [t-0.33]
D = 98.5 t - 32.8333.........Eq2
From Eq1 and Eq2
68.8 t = 98.5 t - 32.83333
29.7 t = 32.83333
t = 1.1
Total time taken(T) = 1.1 hour
Distance = speed × time
Distance = 68.8 × 1.1
Distance = 75.68 km
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
How would the interference pattern produced by a diffraction grating change if the laser light changed from red to blue?
Answer
fringe separation l distance between maxima decreases
Explanation:
Because the wavelength of blue light is smaller than that if red light
Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
After a long walk in the 128C outdoors, a person wearing glasses enters a room at 258C and 55 percent relative humidity. Determine whether the glasses will become fogged.
Answer:
Yes the glasses will be fogged
Explanation:
See attacher file
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%
Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?
Answer:
a) 0.72 V
b) 19.2 mA
c) 2.304 Watts
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = [tex]N_{p}[/tex] = 500 turns
input voltage = [tex]V_{p}[/tex] = 120 V
number of secondary turns = [tex]N_{s}[/tex] = 3 turns
output voltage = [tex]V_{s}[/tex] = ?
using the equation for a transformer
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
substituting values, we have
[tex]\frac{V_{s} }{120 } = \frac{3 }{500} }[/tex]
[tex]500V_{p} = 120*3\\500V_{p} = 360[/tex]
[tex]V_{p}[/tex] = 360/500 = 0.72 V
b) by law of energy conservation,
[tex]I_{P}V_{p} = I_{s}V_{s}[/tex]
where
[tex]I_{p}[/tex] = input current = ?
[tex]I_{s}[/tex] = output voltage = 3.2 A
[tex]V_{s}[/tex] = output voltage = 0.72 V
[tex]V_{p}[/tex] = input voltage = 120 V
substituting values, we have
120[tex]I_{p}[/tex] = 3.2 x 0.72
120[tex]I_{p}[/tex] = 2.304
[tex]I_{p}[/tex] = 2.304/120 = 0.0192 A
= 19.2 mA
c) power input = [tex]I_{p} V_{p}[/tex]
==> 0.0192 x 120 = 2.304 Watts
A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of
40 ut, what magnitude emf is induced between the driver and passenger sides of the truck?
Answer:
The magnitude of the induced Emf is [tex]0.003371V[/tex]
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the [tex]induced EMF= B L v[/tex]
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 28.2 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.25 and \mu_k=\:μ k = 0.12 ). What is the horizontal acceleration of the block?
Answer:
The horizontal acceleration of the block is 4.05 m/s².
Explanation:
The horizontal acceleration can be found as follows:
[tex] F = m \cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]
[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]
Where:
a: is the acceleration
F: is the force exerted by the rope = 28.2 N
θ: is the angle = 30°
[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12
m: is the mass = 5 kg
g: is the gravity = 9.81 m/s²
[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]
Therefore, the horizontal acceleration of the block is 4.05 m/s².
I hope it helps you!
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's magnetic field, the ion will be delfected:
Answer:
East direction
Explanation:
Given that
Charge on the particle is positive.
Moving towards the center of earth .
We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.
Therefore due to the earth magnetic fields , the positive ions will deflect towards East direction.
Thus the answer will be East direction.
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is 64.2° .What is the index of refraction of the reflecting material?
Answer:
n = 2.0686
Explanation:
When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is
n = so tea
let's calculate
n = tan 64.2
n = 2.0686
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the electric field between the two plates?
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
Need help understanding this. If anyone help, that would be greatly appreciated!
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
a) acceleration:
ā=v^2/r
ā=(15m/s)^2/27m
ā=225/27 m/s^2
ā=8.333 m/s^2
force:
F=mā. where the is equal to v^2/r
F=1000kg*8.3 m/s^2
F=8333.3 N
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
upthrust of the liquid on the cube.
Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
define the term DNA
Answer:
It is the carrier of genetic information.
It takes 144 J of work to move 1.9 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates
Answer:
151.58 V
Explanation:
From the question,
The work done in a circuit in moving a charge is given as,
W = 1/2QV..................... Equation 1
Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.
make V the subject of the equation
V = 2W/Q.................. Equation 2
Given: W = 144 J. Q = 1.9 C
Substitute into equation 2
V = 2(144)/1.9
V = 151.58 V
A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
The mutual inductance is [tex]M = 0.000406 \ H[/tex]
Explanation:
From the question we are told that
The number of turns per unit length is [tex]N = 1800[/tex]
The radius is [tex]r = 0.0165 \ m[/tex]
The number of turns of the solenoid is [tex]N_s = 210 \ turns[/tex]
Generally the mutual inductance of the system is mathematically represented as
[tex]M = \mu_o * N * N_s * A[/tex]
Where A is the cross-sectional area of the system which is mathematically represented as
[tex]A = \pi * r^2[/tex]
substituting values
[tex]A = 3.142 * (0.0165)^2[/tex]
[tex]A = 0.0008554 \ m^2[/tex]
also [tex]\mu_o[/tex] is the permeability of free space with the value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]
[tex]M = 0.000406 \ H[/tex]
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 3.62 g coins stacked over the 20.3 cm mark, the stick is found to balance at the 22.5 cm mark. What is the mass of the meter stick
Answer:
0.5792g
Explanation:
The computation of the mass of the meter stick is shown below:
Let us assume the following items
x1 = 50 cm;
m2 = m3 = 3.62 g;
x2 = x3 = 20.3 cm;
xcm = 22.5 cm
Based on the above assumption, now we need to apply the equation of center mass which is given below:
[tex]Xcm = \frac{m1x1 + m2x2 + m3x3}{m1 + m2 + m3} \\\\ 22.5 = \frac{m1\times 50 + 3.62 \times 20.3 + 3.62 \times 20.3}{m1 + 3.62 + 3.62}\\\\ 22.5m1 + 162.9 = 50m1 + 73.486 + 73.486[/tex]
27.5 m1 = 15.928
So, the m1 = 0.5792g
The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what direction must the lens be moved to change the focus of the camera to a person 4.0 m away
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )