Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
Answer:
The velocity is [tex]v = 2.84 1 \ m/s[/tex]
Explanation:
The diagram showing this set up is shown on the first uploaded image (reference Physics website )
From the question we are told that
The mass is m = 4 kg
The length of the string is [tex]L = 2.0 \ m[/tex]
The constant angle is [tex]\theta = 35.4 ^o[/tex]
Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as
[tex]Tcos (\theta ) - mg = 0[/tex]
=> [tex]mg = Tcos (\theta )[/tex]
Now let the force acting on mass horizontally be k so from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{k }{T}[/tex]
=> [tex]k = T sin \theta[/tex]
Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as
[tex]F_v = \frac{m v^2}{r}[/tex]
So
[tex]k = F_v[/tex]
Which
=> [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]
So
[tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]
=> [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]
=> [tex]v = \sqrt{r * g * tan (\theta )}[/tex]
Now the radius is evaluated using SOHCAHTOA rule as
[tex]sin (\theta) = \frac{ r}{L}[/tex]
=> [tex]r = L sin (\theta)[/tex]
substituting values
[tex]r = 2 sin ( 35.4 )[/tex]
[tex]r = 1.1586 \ m[/tex]
So
[tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]
[tex]v = 2.84 1 \ m/s[/tex]
Which of the following statements about Masters programs is not correct?
A. Most Masters athletes did not compete when they were in school.
B. The social life is as important as the athletics on most Masters
teams.
C. The level of competition is not very high in most Masters
programs.
D. Masters programs allow adults to work out and socialize with
people who share their love of a sport.
SUBMIT
The correct answer is C. The level of competition is not very high in most Masters programs.
Explanation:
In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters programs".
An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Answer
0.2067m or 0.2067m
Explanation;
Let lenght of spring= Lo= 11cm=0.110m
It is hang from a mass of
3.05-kg having a length of L1= 12.40cm= 0.124m
Force required to stretch the spring= Fkx
But weight of mass mg= kx then K= Mg/x
K= 3.05-kg× 9.8)/(0.124m-.110m)
K=2135N
But potential Energy U= 0.5Kx
X=√ 2U/k
√(2*10)/2135
X=0.0967m
The required new length= L2= L0 ±x
=
.110m ± 0.0967m
X= 0.2067m or 0.2067m hence the total lenghth
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from the star and is moving at 55 km/s. What is the semimajor axis of the planet's orbit
Answer:
32
Explanation:
Suppose a tank filled with water has a liquid column with a height of 19 meter. If the area is 2 square meters 2m squared, what’s the force of gravity acting on the column of water?
Answer:
372,400 N
Explanation:
The volume of the column is ...
V = Bh = (2 m^2)(19 m) = 38 m^3
If we assume the density is 1000 kg/m^3, then the mass of the water is ...
M = ρV = (1000 kg/m^3)(38 m^3) = 38,000 kg
The force of gravity on that mass is ...
F = Mg = (38,000 kg)(9.8 m/s^2) = 372,400 N
An object has an acceleration of 12.0 m/s/s. If the net force was doubled and the mass were tripled, then the new acceleration would be _____ m/s/s.
✴ Case - I
⟶ Force = F
⟶ Mass = m
⟶ Acceleration = 12m/s²
✴ Case - II
⟶ Force = 2F
⟶ Mass = 3m
To Find :➳ Acceleration in second case.
Concept :⇒ This question is completely based on the concept of newton's second law of motion.
⇒ As per this law, Force is defined as the product of mass and acceleration.
Mathematically, F = ma
Calculation :[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]
New acceleration would be 12 m/s²
Given that;
Acceleration of object = 12 m/s²
New net force = 2f
New mass = 3m
Find:
New acceleration
Computation:
[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]
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A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
radiation transfers energy through___. a metal. b liquid. c touch. d waves.
Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc
Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?
Answer:
Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.
If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2. Determine the orbital period of the satellite.
Answer:
118 minutes( 2 hours approximately )
Explanation:
Here, we are interested in calculating the orbital period of the satellite
Please check attachment for complete solution
Answer:
T = 7101 s = 118.35 mins = 1.9725 hrs
Explanation:
To solve the question, we apply the formula for gravitational acceleration
a = GM/r², where
a = acceleration due to gravity
G = gravitational constant
M = mass of the earth
r = distance between the satellite and center of the earth
Now, if we make r, subject of formula, we have
r = √(GM/a)
Recall also, that
a = v²/r, making v subject of formula
v = √ar
If we substitute the equation of r into it, we have
v =√a * √r
v =√a * √[√(GM/a)]
v = (GM/a)^¼
Again, remember that period,
T = 2πr/v, we already have v and r, allow have to do is substitute them in
T = 2π * √(GM/a) * [1 / (GM/a)^¼]
T = 2π * (GM/a³)^¼
T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼
T = 6.284 * [(3.982*10^14) / 244.140]^¼
T = 6.284 * (1.63*10^12)^¼
T = 6.284 * 1130
T = 7101 s
T = 118.35 mins
T = 1.9725 hrs
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) (1/3), (b) (1/10)
Answer:
35.3°
18.4°
Explanation:
a.
The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.
Using to Law of Malus, I=I0cos²θ
cos²θ=I/I0=(I0/3)/I0/2 ,
cosθ=√2/3−−√=0.6667−−−−−√=0.8165
θ=cos−1(0.8165)=35.3∘
B.
Cos²θ=I/Io =Io/10/Io9
Cosθ= √9/10= 0.9487
= cos−10.9487
=18.4°
(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°
(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°
Angle of polarisation:According to the Malus Law: The intensity of light when passing through a polarizer is given by:
I = I₀cos²θ
where θ is the angle of the polarizer axis with the direction of polarization of the light
I₀ is the initial intensity
When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :
I = I₀<cos²θ> { average of cos²θ, for 0<θ<2π}
I = I₀/2
Now, this emerging light passes through a second polarizer such that:
(a) the intensity is I' = I₀/3
From Malus Law:
I' = Icos²θ
I₀/3 = (I₀/2)cos²θ
cos²θ = 2/3
θ = 35.26°
(b) the intensity is I' = I₀/10
From Malus Law:
I' = Icos²θ
I₀/10 = (I₀/2)cos²θ
cos²θ = 1/5
θ = 63.43°
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g The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
Explanation:
Let the original resistance be R and voltage be V
Applying ohm's law
V / R = 15
V = 15 R
In second case
V / (R+8 ) = 12
V = 12 R + 96
15 R = 12 R + 96
3R = 96
R = 32 ohm .
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]
You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will have a value that is
Answer:
The equivalent is 6R/11Explanation:
We know that the equivalent resistance of resistors connected in parallel is expressed as
[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]
the L.C.M is 6R
[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]
A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answer:
31
Explanation:
Dr. Stein's hypothesis is that excess sugar causes hyperactivity. He is interested in doing research.
Which research method would be the best to use?
Answer:
The correct answer would be - dependent independent variable experiment.
Explanation:
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.
Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, the correct answer would be - dependent independent variable experiment.
The best research method to use for the research of hyperactivity, would be dependent-independent variable experiment.
The given problem is based on the effect of sugar on hyperactivity. Hyper activity refers to the increased movement, impulse actions and a shorter attention span.
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, we can conclude that the best research method to use, would be - dependent-independent variable experiment.
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You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the ratio of the number of turns in the primary to the secondary
Answer:
1:2
Explanation:
It is given that,
Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.
We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.
For a transformer, [tex]\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}[/tex]
So,
[tex]\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}[/tex]
So, the ratio of the number of turns in the primary to the secondary is 1:2.
10 pts! :) If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much work did Kyla do on the bag?
Explanation:
work = force x Distance
w = 10 x 1.5 = 15Nm
The amount of work done by Kyla in lifting the bag is 15 J.
What is meant by work done ?Work done on an object is defined as the cross product of the force applied on the object and the vertical displacement of the object.
Here,
Force applied by Kyla to pick up the bag, F = 10 N
Vertical displacement of the bag, s = 1.5 m
The work done by Kyla in lifting the bag,
W = F x s
W = 10 x 1.5
W = 15 J
Hence,
The amount of work done by Kyla in lifting the bag is 15 J.
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A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s.
Using the formula x = v0 • t + ½ • a • t2, find the time it takes to reach this height? Why do you get two values for time? Explain.
Answer:
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
Explanation:
Given:
initial speed, u = 29 m/s
acceleration due to gravity, g = - 9.8 m/s^2
h = 13 m
Let it is moving with velocity v at a height of 13 m.
Use third equation of motion
v² = u² + 2gh
By substituting the values
v² = 29² - (2 * 9.8 * 13)
v = sqrt 585.94
v = 24.2 m/s
Let it takes time t to reach at height 13 m
Use second equation of motion
s = u * t + 1/2 * g * t²
13 = 29t - 4.9t²
4.9t² - 29t + 13 = 0
using quadratic equation to solve time
29 ± [tex]\sqrt{29^2 - 4 * 4.9 * 13}\\[/tex]
t = ------------------------------------
2 * 4.9
t = 5.43 second or t = 0.49 second
Therefore...
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
An object has an acceleration of 6.0 m/s/s. If the net force was tripled and the mass were doubled, then the new acceleration would be _____ m/s/s.
Answer:
The new acceleration would be 9 m/s².
Explanation:
Acceleration of an object is 6 m/s²
Net force is equal to the product of mass and acceleration i.e.
F = ma
[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]
If the net force was tripled and the mass were doubled, it means,
F' = 3F
m' = 2m
Let a' is new acceleration. So,
[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]
So, the new acceleration would be 9 m/s².
lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2
Answer:
The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]
d = maximum displacement of the mass m₂
1. A coil is formed by winding 250 turns of insulated 16-gauge copper wire, that has a diameter d = 1.3 mm, in a single layer on a cylindrical form of radius 12 cm. What is the resistance of the coil? Neglect the thickness of the insulation and the resistivity of copper is ???? = 1.69 × 10−8 Ω ∙ m.
Answer:
2.39 Ω
Explanation:
Given that
Number of winnings on the coil, = 250 turns
Radius if the copper wire, r(c) = 1.3/2 = 0.65 mm
Radius of single cylinder layer, R = 12 cm
Length of the cylinderical coil, L = 250 * 2π * 12 = 188.4 m
Resistivity of copper, ρ = 1.69*10^-8 Ωm
Area is πr(c)², which is
A = 3.142 * (0.65*10^-3)²
A = 3.142 * 4.225*10^-7
A = 1.33*10^-6 m²
The formula for resistance is given as
R = ρ.L/A, if we substitute, we have
R = (1.69*10^-8 * 188.4) / 1.33*10^-6
R = 3.18*10^-6 / 1.33*10^-6
R = 2.39 Ω.
Therefore, the resistance is 2.39 Ω
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad/s)t].
(a) What is the speed of the wave?
(b) What are the amplitudes of the electric and magnetic fields of this wave?
(c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?
Answer:
a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet