Two waves are represented by y1=Acoswt and y2= Asin (wt-π/6) superimposed on each other. What is the amplitude of the resultant wave?
Displacement will be. [First Method]Y= y1 + y2= Acoswt + Asin(wt- π/6)=Asin(π/2 - wt) +Asin(wt- π/6)= -Asin(wt-π/2) +Asin(wt- π/6)= Asin(wt- π/6) -Asin(wt-π/2)= A(2cos((wt - π/6 + wt - π/2)/2).sin( (wt - π/6 - wt + π/2)/2)=Acos(wt - π/3)So amplitude is A, while phase diff. is π/3Or[Second Method]Y= y1 + y2=Acoswt + A( (Sinwt).(Cos30°) - (Coswt).(Sin30°) )=Acoswt + (A/2).( (Sinwt).sqrt(3) - Coswt )= (A/2)[Coswt + (sqrt(3)).Sinwt ]= A[ (1/2)coswt + ((Sqrt(3))/2).Sinwt ]= A[ cos60°.coswt + sin60°.Sinwt ]= A[cos(wt- 60°)]Or A[ Sin30°coswt + cos30°sinwt] = A[sin(wt +30°) ]So amplitude is A.Ans.
Help with a physics question involving sound waves?
Solution: You are standing at a distance r1 = 85.36 m from a speaker (intensity I1). Standing at the initial distance r1 with an intensity I1: I1 = P1 / A = P1 / (4π×r1²) P1 = I1×4π×r1² Walking to the 25 m closer distance r2 with the double intensity I2. r2 = r1 − 25 I2 = 2×I1 = P2 / (4π×(r1−25)²) P2 =2×I1×4π×(r1−25)² P1 = P2 I1×4π×r1² = 2×I1×4π×(r1−25)² r1² = 2×(r1−25)² "Quadratic equation - WolframAlpha": http://www.wolframalpha.com/input/?i=r%C... The initial distance r1 with the intensity I1 was: r1 = 25×(2±√ (2)) = 85.36 m (The second answer r1 = 14.64 m is not usable, because the distance is too short.) Walking to the 25 m closer distance r2 = 85.36 m − 25 m = 60.36 m. There is the double intensity I2. Help: "How does the sound decrease with distance?" http://www.sengpielaudio.com/calculator-... Use the third calculator and enter: Sound intensity I1 = 1 W/m², reference distance r1 = 85.36 m. Sound intensity I2 = 2 W/m², answer: other distance r2 = 60.36 m. Cheers ebs
The intensity illumination from a light source varies directly as the square of the reciprocal of its distance?
Let the intensity of illumination be i, distance be d given statement indicates i = k* 1/(d^2), where k is an unknown constant Initial & final figures for illumination, distance are i1,d1 & i2,d2 respectively d1= 10 inches, d2 = unknown i2/i1 = 2 => [k*1/(d2^2)]/[k*1/d1^2]=2 => (d1^2)/(d2^2)=2 => d2 = sqrt(d1^2/2) => d2 = sqrt(100/2) = sqrt(50) = 7.07 inches therefore the distance would be 7.07 inches
What will be change in force if distance between charges is increased by 2 times?
The force will go down to 1/4 of what it was. Force between the charges varies inversely to the second power of the distance in between the charges. The equation for the electrical attractive force between two charges is F = (1/4pe)q^2/d^2, where p is pi, e is vacuum permissivity, q is the charge, and d is the distance in between the charges. So if d is doubled F goes down to a 1/4, if d is tripled, F will go down to 1/9, and if d is quadrupled, F will go down to 1/16 and so on… Kaiser T, MD.
Why does bond dissociation enthalpy follow the order Cl2>Br2>F2>I2?
HeyThanks buddy for this question.The Bond dissociation energy of halogens decreases down the group as the size of the halogen is increases .The bond dissociation energy of fluorine is however, lower than chlorine and bromine due to its small size its inter electronic repulsion is very high .That's why the the bond dissociation enthalpy follows the order Cl2 >Br2>F2>I2.Good luckHave a nice day !!AM
Could a maser beam be sent from a satellite?
Literally sending electromagnetic radiation through a building makes it an x-ray, not a microwave laser. Doing so is not difficult. Doing so is dangerous. I do it often, and for imaging purposes as you said.Could we do it from a satellite, ignoring the people would would slowly and painfully kill? No, at least not without some significant advances in engineering technology.A linear accelerator would be too weak to penetrate the atmosphere at such great distances. You’d need a betatron (toroidal accelerator), an incredibly bulky power stabilizer, an entirely new cooling system – conduction works less effectively when the fluid cannot radiate heat through a heat exchange in space, and a source of power. Such a monstrosity would likely be larger than the ISS depending on how the energy is formed (Solar? Nuclear fission? Radiothermal decay?)But before that, how is the satellite orbiting? LEO or GEO - rapid at low altitude or slowly at high altitude? If you want imaging, you need to decide what the subjects are and how to manuever to them.That altitude plays directly into how badly the beam intensity is reduced. Remember that all light abides by the inverse square law. Intensity is directly inversely proportion to the square of the change in distance. E.g., intensity value of 100 at 1 meter = intensity of 25 at 2 meters.So you’ve got essentially a death ray putting out 30 million electronvolt energies at 15,000 RADs per minute at one meter? That’s an impressive machine. 30 MEV is really something. That power transformer and cooling engineering feat is no joke.Let’s math it. I1 • (D1^2)=I2 • (D2^2) where you’re solving for Intensity2 with known Distances 1 & 2, and known Intensity1.15,000 RADs • (1 meter squared) = Unknown • (450 kilometers at Low Earth Orbit, squared)15,000 • 1^2 = 15,000 because 1 squared is 1.450 km is 450,000 meters. Squared that is 202,500,000,000.That gives us 15,000 = Unknown • 202,500,000,000Divide both sides by 202,500,000,000 to make I2 solitary.That gives us Intensity of 0.000000074074 RADs per minute at 450 kilometers. Or 38.9 milliRAD per year. The LD50 is 500 RADs. The public allowance is 2 whole RADs per year.That’s not much of a death ray, bro. The guy who gets sent up there to adjust the cooling hose gets more radiation from the Sun in one trip than anyone on Earth gets from the machine in a whole year.
HELP Chemistry homework?
A) At a distance of 41 feet, the radiation source delivers 4.5 rem of radiation. The dose delivered in the period of time in question is inversely proportional to 1/r^2. So if you cut the distance in half, you increase the dose by 2 squared. If you cut it down to a third, the dose is increased by a factor of 3 squared. We want to keep the dose at or below 25 rem. This means we can increase the dose by a factor of up to 25/4.5 = 5.55 = 50/9 D = k/r^2 If we call the different doses D1 and D2 at distances r1 and r2, then the relation between the doses and the distances is: D1/D2 = r2^2/r1^2 So 25/4.5 = 41^2/r1^2 Solve for r1: 4.5/25 = r1^2/41^2 0.18 = r1^2/1681 r1^2 = 0.18*1681 = 302.58 r1 = 17.39 feet Checking: 17.39^2 / 41^2 = 302.41 / 1681 = 0.18 Just what we wanted it to be. B) A solution's activity will not be measured in mL. It may be measured in Ci/mL or rem/mL, but it has to be in the form of units of radioactivity divided by volume so that when you multiply by the volume used you're left with units of radioactivity. If you have a solution that delivers 165 mrem/mL and inject 5 mL into a patient, the patient receives a dose of 165 mrem/mL * 5 mL = 825 mrem * mL/mL = 825 mrem.
CALC HELP!! please. Chapter 3 From Calculus nth edition (application of differentiation)?
Imagine a number line with Light Source 1 (L1) at x = 0, and Light Source 2 (L2) at x = d, where the positive x direction is from L1 to L2. Then you can express the distances from L1 and L2 (call these distances d1 and d2) to any point on the number line with respect to d and x. Namely: d1 = x d2 = d - x The following relationship between intensity distance and illumination was given: Q = I / D² (where Q is illumination, I is intensity, and D is distance). Naturally, the total illumination that any point on our number line will receive is going to be the sum of the illumination from either light source, so: Q = (I1) / (d1)² + (I2) / (d2)² = (I1) / x² + (I2) / (d - x)² This describes the total illumination of every point on the number line relative to position x. We need to minimize this function, so take its derivative: Q' = -2*(l1) / x³ + 2*(l2) / (d - x)³ Set this equal to zero and solve for x to locate critical points: 0 = -2*(l1) / x³ + 2*(l2) / (d - x)³ (l1) / x³ = (l2) / (d - x)³ ³√[ (l2) / (l1) ] = (d - x)/x x = d / ( ³√[ (l2) / (l1) ] + 1 )