Answer:
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Explanation:
1. FALSE
The acceleration at the top point is equal to the acceleration due to gravity (9.8 m/s²). It is the velocity of the stone that becomes zero for a moment, at the top point.
2. FALSE
The acceleration of the body in a free fall motion always remains constant, and its value is equal to the acceleration due to gravity (9.8 m/s²)
3. FALSE
The velocity and acceleration point in opposite direction, during upward motion only (Velocity points up and acceleration points down). But they point in same direction during the downward motion (Both velocity and acceleration point down).
4. TRUE
Since, there is no air friction. Therefore, the acceleration for both upward an downward motion will be constant, without any opposing force. Hence, the time taken by the stone to go up will be equal to time taken by the stone to come down.
A heater rod (10 mm diameter, 100 mm length) of emissivity 0.75 is enclosed within a hollow cylindrical vacuum chamber (50 mm diameter, 100 mm length) of emissivity 0.25. The entire setup is insulated at the top and bottom ends by a low emissivity material, preventing any conductive heat dissipation from the ends. The heater rod is known to have a surface temperature of 1000 K, while the vacuum chamber is at a surface temperature of 300 K. How much heat is dissipated from the heater rod to the vacuum chamber (W)
Answer:
Explanation:
Given that:
Heater temperature ,T₁ = 1000K
Vaccum Chamber ,T₂ = 300K
emissivity of heater E₁ = 0.75
emissivity vaccum E₂ = 0.25
Heater diameter d₁ = 10 * 10⁻³mm
vaccum chamber d₂ = 50 * 10⁻³mm
When there is vaccum, then no air resistance will be there,
F₁₂ = 1
F₁₁ = 0
[tex]R_1= \frac{1-E_1}{E_1A_1} \\\\=\frac{1-0.75}{0.75*\pi * 10^-^2*L}[/tex]
[tex]R_2=\frac{1}{F_1_2 * A_1} \\\\=\frac{1}{1* \pi *10^-^2*L}[/tex]
[tex]R_3=\frac{1-0.25}{F_1_2 * A_1} \\\\=\frac{1}{0.25* \pi *5*10^-^2*L}[/tex]
Heat leaving from heater surface 1 to vaccum
[tex]Q_1_2 = \frac{L \pi \sigma (T_1^4- T_2^4)}{R_1+R_2+R_3}[/tex]
[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)}{\frac{0.25}{0.75*10^-2}+\frac{1}{10^-2} +\frac{0.75}{0.25*10^-^2*5} }[/tex]
[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)} {0.3333+1+0.6}\\\\Q_1_2= 91.39 \text {watt}[/tex]
How does the centripetal force depend on speed, radius and the revolving mass?
Answer:
Mass velocity and radius are all related to centripetal force
Explanation:
By frequency of its rotation and the radius of the circular path along which objects moves
Which of these is not a factor that influences animal growth? *
2 points
A. inherited diseases
B. gravitational force
C. genetic information
D. availability of food
What does it mean that an exoplanet exists in the "habitable zone"? Question 6 options: (a) It is located the same distance from its star that Earth is from the Sun. (b) It is located the right distance from its star to enable liquid water to exist on its surface. (c) It means that the exoplanet has an oxygen rich atmosphere. (d) It is located a distance from its star that makes extinction level asteroid impacts unlikely.
Answer:
(b) It is located the right distance from its star to enable liquid water to exist on its surface.
Explanation:
The habitable zone according to astronomy and astrobiology is a region around a star that the planets around the star can hold and support liquid water. For our solar system, the habitable zone coincides approximately with the distance from the sun to Earth.
1.How do we define electric potential at any point in space ?
2.How does the electric potential change along the direction of the net electric field?Explain them in details.
Answer:
1. The electric potential at any point in space, is the work done in moving a unit charge from infinity to that point.
2.The electric potential varies radially by decreasing as the distance from the electric field increases.
Explanation:
1. The electric potential at any point in space, is the work done in moving a unit charge from infinity to that point. It is defined mathematically as, V = W/q where V = electric potential, W = work done in moving charge q into electric field and q = charge. For the work done of moving a charge q into an electric field of charge q' at a distance r between them, W = kqq'/r and work done per unit charge in moving q in electric field is V = W/q = kqq'/r ÷q = kq'/r, Also, for an electric field, E, V = ∫E.dr where r is the direction moved by the charge in the electric field.
2. Since the electric potential V = kq'/r, V ∝ 1/r since k and q' are constant. So, the electric potential varies radially by decreasing as the distance from the electric field increases.
Where is the potential energy equal to zero?
Answer:
im sure your already past this but it's E.
Explanation:
This is because in this case potential energy is linear to height, which means that the higher the more potential energy.
Need help on this question I’d really appreciate it thanks!
Answer:radiation
Explanation:
radiation is the only one that makes sense
Some animals, such as dolphins, use ________ to navigate and find food. This method uses sound waves.
Suppose multiple stations are connected in wired network where CSMA/CD implemented. Among them, station A wants to communicate with B and station C wants to communicate with D. Previously, both of these sender stations have faced collisions at different timings and are still unsuccessful in their transmission. The value of k for station A has reached to 6, whereas the value of k for station B has reached to 4. The maximum attempts allowed to any station is set to 10 and transmission time for a frame is 0.5ms.
This time, Both of the sender stations (A & C) sensed the medium, both got the channel idle and started transmission. Sooner, the station A detected the collision.
i.What will happen now?
ii.How long station A has wait before re-trying
iii.Meanwhile what will happen at station C
Answer:
bro i think ur from the uni of arid agriculture rawalpindi and today is your networking paper
Explanation:
The displacement of the air molecules in a sound wave is modeled with the wave function s(x, t) = 3.00 nm cos(50.00 m−1x − 1.71 ✕ 104 s−1t). (a) What is the wave speed (in m/s) of the sound wave? 342 Correct: Your answer is correct. m/s (b) What is the maximum speed (in m/s) of the air molecules as they oscillate in simple harmonic motion? m/s (c) What is the magnitude of the maximum acceleration (in m/s2) of the air molecules as they oscillate in simple harmonic motion?
Answer:
a) 342 m/s
b) 51*10^-6 m/s
c) 0.87m/s^2
Explanation:
The following function describes the displacement of the molecules in a sound wave:
[tex]s(x,t)=3.00nm\ cos(50.00\ m^{-1}x-1.71*10^4s^{-1}t)[/tex] (1)
The general form of a function that describes the same situation is:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
By comparing equations (1) and (2) you have:
k: wave number = 50.00 m^-1
w: angular frequency = 1.71*10^4 s^-1
A: amplitude of the oscillation = 3.00nm
a) The speed of the sound is obtained by using the formula:
[tex]v=\frac{\omega}{k}=\frac{1.71*10^4s^-1}{50.00m^{-1}}=342\frac{m}{s}[/tex]
b) The maximum speed of the molecules is the maximum value of the derivative of s(x,t), in time. Then, you first obtain the derivative:
[tex]\frac{ds}{st}=-\omega A sin(kx-\omega t)[/tex]
The max value is:
[tex]v_{max}=\omega A[/tex]
[tex]v_{max}=(1.71*10^4s^-1)(3.00nm)=51300\frac{nm}{s}=51\frac{\mu m}{s}[/tex] = 51*10^-6 m/s
c) The acceleration is the max value of the derivative of the speed, that is, the second derivative of the displacement s(x,t):
[tex]a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=-\omega^2A cos(kx-\omega t)\\\\a_{max}=\omega^2 A[/tex]
Then, the maximum acceleration is:
[tex]a_{max}=(1.71*10^4s^{-1})^2(3.00nm)=0.87\frac{m}{s^2}[/tex]
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?PLEASE SORT QUESTION BELOWSort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.mQ: the mass of the quarterbackmB: the mass of the football(vQx)i: the horizontal velocity of quarterback before throwing the ball(vBx)i: the horizontal velocity of football before being thrown(vQx)f: the horizontal velocity of quarterback after throwing the ball(vBx)f: the horizontal velocity of football after being thrown
Answer:
a
The speed of the quarterback backward is [tex]v_q = 0.08 \ m/s[/tex]
b
Known are
[tex]m_Q , m_B , (v_{Bx})_i (v_{Qx})_f, (v_{Bx})_f[/tex]
Unknown
[tex](v_{Qx})_f[/tex]
Explanation:
From the question we are told that
The mass of the quarterback is [tex]m_Q = 80 \ kg[/tex]
The mass of the ball is [tex]m_B = 0.43 \ kg[/tex]
The speed of the ball is [tex]v_{B x}= 15 \ m/s[/tex]
The law of momentum conservation can be mathematically represented as
[tex]m_Q u_{Qx} + m_Bu_{Bx} = - m_{Q} v_{Qx} + m_B v_{Bx}[/tex]
Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball
So
[tex]m_Q v_{Qx} = m_B v_ {Bx}[/tex]
=> [tex]v_{Qx} = \frac{m_Bv_{Bx}}{m_Q}[/tex]
substituting values
[tex]v_q = \frac{0.43 * 15}{80}[/tex]
[tex]v_q = 0.08 \ m/s[/tex]
What is the acceleration of an object that takes 20 sec to change from a speed of 200 m/s to 300 m/s ?
Given:
initial velocity, u = 200 m/sFinal velocity, v = 300 m/s Time taken, t = 20 secTo be calculated:
Calculate the acceleration of given object ?
Formula used:
Acceleration = v - u / t
Solution:
We know that,
Acceleration = v - u / t
☆ Substituting the values in the above formula,we get
Acceleration ⇒ 300 - 200 / 20
⇒ 100/20
⇒ 5 m/s²
A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?
Answer:
E = 0.5N/C
Explanation:
In order to calculate the magnitude of the electric field you use the following formula:
[tex]E=\frac{F}{q}[/tex]
q: charge = 5.0*10^-4 C
F: force on the charge = 2.5*10^-4N
You replace the values of q and F in the equation for E:
[tex]E=\frac{2.5*10^{-4}N}{5.0*10^{-4}C}=0.5\frac{N}{C}[/tex]
hence, the magnitude of the electric field at the position of the carge is 0.5N/C
At the location of the test charge, the magnitude will be:
"0.5 N/C".
Electric fieldWhenever charge seems to be available inside any form, an electric property has been linked among each spatial position. This same size and direction are represented either by the quantity of E.
According to the question,
Charge, q = 5.0 × 10⁻⁴ C
Force on the charge, F = 2.5 × 10⁻⁴ N
We know the relation of the magnitude of electric field be
→ E = [tex]\frac{F}{q}[/tex]
By substituting the values, we get
= [tex]\frac{2.5\times 10^{-4}}{5.0\times 10^{-4}}[/tex]
= 0.5 N/C
Thus the approach above is correct.
Find out more information about electric field here:
https://brainly.com/question/14372859
A spherical conductor with a 0.103 m radius is initially uncharged. How many electrons should be removed from the sphere in order for it to have an electrical potential of 4.50 kV at the surface
Answer:
Explanation:
capacitance of the capacitor C = 4πε₀R
= 1 x .103 / (9 x 10⁹ ) [ 1/ 4πε₀ = 9 x 10⁹ . ]
= .0114 x 10⁻⁹ F
potential V = 4.5 x 10³ v
charge Q = CV
.0114 x 10⁻⁹ x 4.5 x 10³
= .0515 x 10⁻⁶ C
charge on one electron = 1.6 x 10⁻¹⁹
no of electrons to be removed
= .0515 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .032 x 10¹³
3.2 x 10¹¹ electrons.
A 4 kW vacuum cleaner is powered by an electric motor whose efficiency is 90%. (Note that the electric motor delivers 4 W of net mechanical power to the fan of the cleaner). What is rate at which this vacuum cleaner supplies energy to the room when running
Answer:[tex]3.6\ kW[/tex]
Explanation:
Given
Power Supplied [tex][tex]P_{input}=4\ kW[/tex][/tex]
Efficiency of the motor [tex]\neta =90\%[/tex]
and [tex]\neta =\dfrac{\text{Power output}}{\text{Power input}}[/tex]
[tex]\Rightarrow 0.9=\dfrac{P_{out}}{4}[/tex]
[tex]\Rightarrow P=0.9\times 4[/tex]
[tex]\Rightarrow P=3.6\ kW[/tex]
So, vacuum cleaner delivers a power of [tex]3.6\ kW[/tex]
An earth observer sees an alien ship pass overhead at 0.3c. The ion gun of the alien ship shoots ions straight ahead of the ship at a speed 0.4c relative to the ship. What is the speed of the ions relative to the earth observer?
Answer:
The velocity of the ions relative to the earth observer is [tex]v = 0.63 c[/tex]
Explanation:
From the question we are told that
The speed of the alien ship is [tex]v_a = 0.3c[/tex]
The speed of the iron gun is [tex]v_g = 0.4c[/tex]
The motion of the ions relative to the earths observer can be mathematically represented as
[tex]v = \frac{v_a + v_g}{[\frac{1+ v_av_g}{c^2} ]}[/tex]
substituting values
[tex]v = \frac{ 0.3c + 0.4c}{[\frac{1+ 0.3c * 0.4c}{c^2} ]}[/tex]
=> [tex]v = 0.63 c[/tex]
A substance is boiled repeatedly and stirred, but the solute never mixes with the solvent. Which best describes why? The temperature was not high enough to mix the solute and solvent. The chemical properties of the solute and solvent are different. The pressure was increased. The solution was saturated.
Answer:
b
Explanation:
If a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent then the chemical properties of the solute and solvent are different, therefore the correct answer is option B.
What is a Chemical compound?A chemical compound is a combination of two or more either similar or dissimilar chemical elements.
For example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.
As given in the problem a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent,
The correct response is option B because the chemical characteristics of the solute and solvent are different if the material is repeatedly boiled and stirred without the solute and solvent mixing.
To learn more about a chemical compound, refer to the link ;
brainly.com/question/12166462
#SPJ6
Describe how heat transfer occurs when you
place your hand in a sink full of hot water.
Answer:
If one places ones's hands into a sink full of hot water, heat is transferred from the hot water to the hands. Heat only flows when there is a temperature gradient, i.e when there is a difference in temperature between a hot and a cold body. Heat is always transferred from a hot body to a cold body and never the reverse in a normal system.
At the boundary of the hand and water, heat transfer is by conduction between the hot water molecules and the molecules of the hands. This heat is then further transferred to the internal body organs and body fluid through either convection or conduction. This heat transfer is maintained until both the water and the hands are at the same temperature, assuming the owner is able to withstand the heat.
Learning Goal: To understand the cause of constructive and destructive interference for the double-slit experiment, and to explain how the interference pattern depends on the parameters of the emitted waves. For this tutorial, use the PhET simulation Wave Interference. This simulation allows you to send waves through a variety of barriers and look at the resulting interference patterns.
Answer:
a) it is essential that the waves lurk coherently
b)the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one,
Explanation:
In double slit interference experiments it is essential that the waves lurk coherently, that is, that the relative phase of the waves that reaches each slit is maintained over time, this is achieved when point sources are used by passing light through a initial slit and if a laser is used it is already consistent at the output.
When the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one, when this path difference is equal to a whole number of donut lengths, it has constructive interference and if it is a number sowing we have destructive interference.
d sin θ = m λ constructive
d sin θ = (m + ½) λ destructive
where d is the distance between the two slits, lam is the wavelength and m is an integer called the interference order
Beats are the result of
Answer:
The phenomenon of beats is the result of sound interference and sound diffraction in periodic vibrations. In periodic vibration, the beats of sounds produced by the interference and diffraction of the waves of different frequencies.
Explanation:
Beats. When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats.
4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3
vectors:
TĀ = 4.5N, 0,= 55
BI = 1.5N, 0,= 145
C = 6.00, 0 = 235
Determine the resultant vector Ř of these 3 vectors through the analytic method
involving components. You have state both the magnitude and the angle with respect to
x-axis.
Expand each vector into their component forms:
[tex]\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N[/tex]
Similarly,
[tex]\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N[/tex]
[tex]\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N[/tex]
Then assuming the resultant vector [tex]\vec R[/tex] is the sum of these three vectors, we have
[tex]\vec R=\vec A+\vec B+\vec C[/tex]
[tex]\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N[/tex]
and so [tex]\vec R[/tex] has magnitude
[tex]\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N[/tex]
and direction [tex]\theta_R[/tex] such that
[tex]\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ[/tex]
A constant-velocity horizontal water jet from a stationary nozzle impinges normally on a vertical flat plate that rides on a nearly frictionless track. As the water jet hits the plate, it begins to move due to the water force. As a result, the acceleration will _____.
Answer:
a = ½ ρ A/M v₁²
Explanation:
This is a problem of fluid mechanics, where the jet of water at constant speed collides with a paddle, in this collision the water remains at rest, we write the Bernoulli equation, we will use index 1 for the jet before the collision the index c2 for after the crash
P₁ + ½ ρ v₁² + ρ g h₁ = P₂ + ½ ρ v₂² + ρ g h₂
in this case the water remains at rest after the shock, so v₂ = 0, as well as it goes horizontally h₁ = h₂
P₁-P₂ = ½ ρ v₁²
ΔP = ½ ρ v₁²
let's use the definition of pressure as a force on the area
F / A = ½ ρ v₁²
F = 1/2 ρ A v₁²
the density is
ρ = m / V
the volume is
V = A l
F = ½ m / l v₁²
knowing the force we can focus on the acceleration of the mass palette M
F = M a
a = F / M
a = ½ m/M 1/l v₁²
as well it can be given depending on the density of the water
a = ½ ρ A/M v₁²
5. Infer What might happen to sand in a
valley if more sand is deposited on top of it?
Answer:
It may turn into sedimentary rock when exposed to pressure.
Explanation:
When successive layer of sand is deposited on top of sand, the bottom layers go deeper and deeper. They are exposed to high pressures and temperatures and turn into hard rocks.
Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric force on this charged particle A depend:_____.
1. the size and shape of the point charge Q
2. the specific location of the point charge Q (while the distance between Q and A is fixed)
3. the mass of the charged particle A
4. the size and shape of the charged particle A
5. the distance between the point charge Q and the charged particle A
6. the type of charge on the charged particle A
7. the mass of the point charge Q
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
10. the specific location of the charged particle A (while the distance between Q and A is fixed)
11. the relative orientation between Q and A (while the distance between Q and A is fixed)
Answer:
Explanation:
The magnitude of the electric force on this charged particle A depends upon the following
5. the distance between the point charge Q and the charged particle A
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
What is kinetic energy
Answer:
Kinetic energy is the energy of motion
Answer:
Energy which a body possesses by virtue of being in motion
Explanation:
A rock with mass m = 3.60 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 15.0 N (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force f=kv, where v is the speed in m/s and k = 2.68 N×s/m.(a) Find the initial acceleration a0 (b) Find the acceleration when the speed is 3.60 m/s.
Answer:
1
Explanation:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
What is the effect on the period of a pendulum if you decrease its length by 6.35%? (Answer this question in terms of the initial period T.) T' = 0.87703 Incorrect: Your answer is incorrect. T
Answer:
T' = 0.9677T
Explanation:
The period of a pendulum is given by the following formula:
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
l: length of the pendulum
g: gravitational acceleration
If the length of the pendulum is decreased in 6.35% the length of the pendulum becomes:
[tex]l'=l-0.0635l=0.9365l[/tex]
The new period for a length of l' is:
[tex]T'=2\pi \sqrt{\frac{l'}{g}}=2\pi \sqrt{\frac{0.9365l}{g}}=\sqrt{0.9365}(2\pi \sqrt{\frac{l}{g}})=0.9677(2\pi \sqrt{\frac{l}{g}})\\\\T'=0.9677T[/tex]
hence, the new period is 0.9677T
The planet Earth travels in a circular orbit at constant speed around the Sun. What is the net work done on the Earth by the gravitational attraction between it and the Sun in one complete orbit
Answer:
WT = 3.32*10^34 J
Explanation:
The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:
[tex]W_T=\int F_g dr[/tex] (1)
Fg: gravitational force between Sun and Earth
The gravitational force is given by:
[tex]F_g=G\frac{m_sm_e}{r^2}[/tex] (2)
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
ms: mass of the sun = 1.989*10^30 kg
me: mass of the Earth = 5.972 × 10^24 kg
r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km
You replace the formula (2) in (1):
[tex]W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}[/tex]
Next, you replace the values of all variables and solve obtain WT:
[tex]W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J[/tex]
hence, the work done on the Earth, in one orbit, is 3.32*10^34 J
A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They computed their average speed for the trip to be 55.7 mi/h.How many hours did the trip take?
Answer:
Time taken for trip = 12.74 hour (Approx)
Explanation:
Given:
Distance of trip = 710-mi
Average speed for the trip = 55.7 mi/h
Find:
Time taken for trip = ?
Computation:
⇒ Time = Distance / Speed
⇒ Time taken for trip = Distance of trip / Average speed for the trip
⇒ Time taken for trip = 710-mi / 55.7 mi/h
⇒ Time taken for trip = 12.74 hour (Approx)
The amplitudes and phase differences for four pairs of waves of equal wavelengths are (a) 2 mm, 6 mm, and π rad; (b) 3 mm, 5 mm, and π rad; (c) 7 mm, 9 mm, and π rad; (d) 2 mm, 2 mm, and 0 rad. Each pair travels in the same direction along the same string. Without written calculation, rank the four pairs according to the amplitude of their resultant wave, greatest first. (Hint: Construct phasor diagrams.)
Answer:
a = d > b = c
Explanation:
The information about amplitudes and phase differences for four pairs of waves of equal wavelengths are given below:
(a) 2 mm, 6 mm, and π rad
(b) 3 mm, 5 mm, and π rad
(c) 7 mm, 9 mm, and π rad
(d) 2 mm, 2 mm, and 0 rad
Whenever a wave has zero phase difference, its amplitude of the resultant wave will be twice the amplitude of any of the two waves. Nevertheless, let assume that the amplitude is a vector having angle Ø between them. The resultant vector will help us rank the four pairs according to the amplitude of their resultant wave by using phasor diagrams.
a.) 6 - 2 = 4mm
b.) 5 - 3 = 2mm
c.) 9 - 7 = 2mm
d.) 2 + 2 = 4mm
Therefore,
a = d and b = c
a = d > b = c
Please find the attached file for the phasor diagrams
