Answer and Explanation:
When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.
Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.
Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:
P = IV -------------(i)
Where;
P = Power rating = 1000W
I = Current used
V = Voltage rating = 240V
Substituting these values into equation (i) gives;
1000 = I x 240
I = [tex]\frac{1000}{240}[/tex] = 4.17A
Therefore, the current used by the stove is 4.17A
To get the resistance R of the stove, we can use the relation;
P = [tex]\frac{V^2}{R}[/tex]
R = [tex]\frac{V^2}{P}[/tex]
R = [tex]\frac{240^2}{1000}[/tex]
R = 57.6Ω
Therefore, the resistance of the stove is 57.6Ω
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:
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.
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:
Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
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
Exercise 1 - Questions 1. Hold the grating several inches from your face, at an angle. Look at the grating that you will be using. Record what details you see at the grating surface. 0 Words 2. Hold the diffraction grating up to your eye and look through it. Record what you see. Be specific. 0 Words 3. Before mounting the diffraction grating, look through the opening that you made for your grating. Record what you see across the back of your spectroscope.
Answer:
1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
2)If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum
3) must be able to see the well-collimated light emission source
Explanation:
1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.
These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.
The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.
2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.
If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.
3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.
The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.
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.
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:
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
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 magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temperature of 850°C, is floating in space, rotating about its axis with an angular speed of 20.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
Answer:
A. 9.31 x10^10J
B. -8.47x10 ^ 12J
C. 8.38x 10^12J
Explanation:
See attached file pls
a car brakes and stops at 10 [m]. While stopping, the friction force of the wheels on the pavement is 400 [N]. Calculate the work done.
Explanation:
Work = force × distance
W = (400 N) (10 m)
W = 4000 J
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
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above
the horizontal, Figure 1. A constant force of 18 N is applied on the box in a direction 10°
with the horizontal causing the box to accelerate up the incline.
The coefficient of
kinetic friction between the block and the plane is 0.25.
Show the free body diagrams
(a) Calculate the block's
acceleration as it moves up the incline. (6 marks)
(b) If the block slides down at a constant speed, find the value of force applied.
(4 marks)
Answer:
a) a = 17.1 m / s², b) F = 3.04 N
Explanation:
This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities
* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components
* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components
We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is
θ = 10 -30 = -20º
The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force
sin (-20) = F_{y} / F
cos (-20) = Fₓ / F
F_{y} = F sin (-20)
Fₓ = F cos (-20)
F_y = 18 sin (-20) = -6.16 N
Fₓ = 18 cos (-20) = 16.9 N
The decomposition of the weight is the customary
sin 30 = Wₓ / W
cos 30 = W_y / W
Wₓ = W sin 30 = mg sin 30
W_y = W cos 30 = m g cos 30
Wₓ = 0.8 9.8 sin 30 = 3.92 N
W_y = 0.8 9.8 cos 30 = 6.79 N
Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane
Now we can write Newton's second law for each axis
X axis
Fₓ - fr = m a
Y Axis
N - [tex]F_{y}[/tex] - Wy = 0
N =F_{y} + Wy
N = 6.16 + 6.79
They both go to the negative side of the axis and
N = 12.95 N
The friction force has the formula
fr = μ N
we substitute
Fₓ - μ N = m a
a = (Fₓ - μ N) / m
we calculate
a = (16.9 - 0.25 12.95) / 0.8
a = 17.1 m / s²
b) now the block slides down with constant speed, therefore the acceleration is zero
ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced
Newton's law for the x axis
Fₓ -fr = 0
Fₓ = fr
F cos 20 = μ N
F = μ N / cos 20
we calculate
F = 0.25 12.95 / cos 20
F = 3.04 N
this is the force applied at an angle of 10º to the horizontal
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The rod is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
angular frequency ω = 2π / T
= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8
Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,
Then,
Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]
Real depth = Index of refraction x apparent depth
Since the same container is used, we can make an assumption that;
real depth of fluid 1 = real depth of fluid 2
That is,
1.37 x 9 = n x 6.86
Where n = Index of refraction for the second fluid.
make n the subject of formula
n = 12.33 / 6.86
n = 1.79
Therefore, the index of refraction of the second fluid is 1.8 approximately.
Learn more about refraction here: https://brainly.com/question/10729741
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
The _____________ is the thermonuclear fusion of hydrogen to form helium operating in the cores of massive stars on the main sequence
Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?
Answer:
The wavelength is [tex]\lambda = 1029 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the left light is [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]
The voltage across A and B is [tex]V_{AB } = 1.2775 \ V[/tex]
Let the stopping potential at A be [tex]V_A[/tex] and the electric potential at B be [tex]V_B[/tex]
The voltage across A and B is mathematically represented as
[tex]V_{AB} = V_A - V_B[/tex]
Now According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side in terms of electron volt is mathematically represented as
[tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]
Where W is the work function of the metal
h is the Planck constant with values [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
And the stopping potential at B for the ejected electron from the right side in terms of electron volt is mathematically represented as
[tex]eV_B = \frac{h * c}{\lambda } - W[/tex]
So
[tex]eV_{AB} = eV_A - eV_B[/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]
=> [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]
=> [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]
Where e is the charge on an electron with the value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]
=> [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]
=> [tex]\lambda = 1.029 *10^{-6} \ m[/tex]
=> [tex]\lambda = 1029 nm[/tex]
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
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
A brick of mass M has been placed on a rubber cushion of mass m. Together they are sliding to the right at constant velocity on an ice-covered parking lot. (a) Draw a free-body diagram of the brick and identify each force acting on it. (b) Draw a free-body diagram of the cushion and identify each force acting on it. (c) Identify all of the action–reaction pairs of forces in the brick–cushion–planet system.
A) The free-body diagram of the forces acting on the brick is attached.
B) The free-body diagram of the forces acting on the rubber cushion is attached.
C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.
A) The brick has a Mass M placed on top of a rubber cushion of mass m.
This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;
Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Gravitational force acting on brick = Mg
Find attached the free body diagram.
B) The forces acting on the cushion will be;
Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Gravitational force of earth acting on cushion = mg
Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
C) The action pairs of forces are;
i) Force; Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
ii) Action Force; Gravitational force acting on brick = Mg
Reaction; Gravitational force of brick acting on the earth
iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Reaction; Force of rubber cushion on parking lot pavement
iv) Action Force; Gravitational force of earth acting on rubber cushion = mg
Reaction Force; Gravitational force of rubber cushion on the earth.
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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)
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.
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?
Answer:
(a) 0.59 ms
(b) 0.15 ms
(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.
Explanation:
(a) distance between ears = 20 cm = 0.2 m
speed of sound generated = 340 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{340}[/tex]
= 5.8824 × [tex]10^{-4}[/tex]
= 0.59 ms
The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.
(b) distance between ears = 20 cm = 0.2 m
speed of sound in water = 1530 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{1530}[/tex]
= 1.4815 × [tex]10^{-4}[/tex]
= 0.15 ms
The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.
(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.
A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds
B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds
C) The significance about the time interval of the two situations is that;
Transmission of sound varies with different mediums.
A) We are given;
Distance between the two ears; d = 20 cm = 0.2 m
Speed of sound; v = 340 m/s
Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;
t = d/v
t = 0.2/340
t = 0.588 × 10⁻³ seconds
B) We are now told that the speed of sound in water is 1530 m/s. Thus;
t = 0.2/1530
t = 0.131 × 10⁻⁵ seconds
C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.
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Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=
Answer:
Electric potential at position, x = -20 m, = -100 V
Electric potential at position, x = 0 m, = 0
Electric potential at position, x = 10 m, = 50 V
Electric potential at position, x = 11 m, = 55 V
Electric potential at position, x = 99 m, 495 V
Explanation:
Given;
magnitude of electric field, E = 5.0 V/m
at position x = 10 m, electric potential = 50 V
Electric potential at position, x = -20 m
V = Ex
V = 5 (-20)
V = -100 V
Electric potential at position, x = 0 m
V = Ex
V = 5(0)
V = 0
Electric potential at position, x = 10 m
V = Ex
V = 5(10)
V = 50 V
Electric potential at position, x = 11 m
V = Ex
V = 5(11)
V = 55 V
Electric potential at position, x = 99 m
V = Ex
V = 5(99)
V = 495 V