What is the unit vector perpendicular to the following vectors: [math]2 \hat \imath + 2 \hat \jmath - \hat k[/math] and [math]6 \hat \imath -3 \hat \jmath +2 \hat k[/math]?
Answer is [math]\;\;\pm\big([/math][math]\frac{ i +8j +18k}{\sqrt{389}}\big) \;\;[/math]Method:Suppose the vectorai +bj+k is perpendicular to both vectors [math]\overline{u} = 2i+2j-k [/math] and[math]\overline{v} = 6i - 3j+k[/math]By perpendicularity, we get[math]2a + 2b - 1 =0 [/math] and[math]6a - 3b +1 = 0[/math]Solving these equations we get[math]a=1/18 , b= 4/9[/math]Therefore the unit vector in the direction of [math]18 (ai+bj+1k) = i + 8j + 18k[/math][math]( ie \;\;\frac{ i +8j +18k}{\sqrt{389}}) [/math] is a vector of the required type.[math]\;\;-(\frac{ i +8j +18k}{\sqrt{389}}) [/math] is also a vector of the required type.Another method:[math]\overline{u}\times [/math][math]\overline{v} =(2i+2j-k) \times( 6i - 3j+k) \;[/math][math]\;\;\;=\;(2-3) i+(-6-2) j+(-6-12) k\;=\;-1i-8j-18k[/math]Hence the unit vector in this direction is[math] \;\;-\big(\frac{ i +8j +18k}{\sqrt{389}}\big). [/math]Hence both vectors[math]\;\;\pm(\frac{ i +8j +18k}{\sqrt{389}}) [/math] are unit vectors perpendicular to the given vectors.
What would be the duration of the year if the distance between the earth and the sun was doubled?
if earth gets double far away from sun as now than sinceR=distance of earth from sun(current)R’=distance of earth from sun(when doubled) and hence,R’=2RT=time period of revolution of earth when it is at R distance = 24hrsT’= time period of distance when earth is at R’ distanceNow,(T^2) / (T’ ^2) = (nR^3) / (nR’ )^3 (Keplar’s law)(here n is proportionallity constant)thereby(24^2) / (T’ ^2) = (R^3) / (2R )^3(24^2) * 8 = T’ ^2{24 * [2*2^(1/2)]}^2=T’ ^248 * 2^(1/2)=T’Hence time period will be increased by 24 * 2^(1/2) times
Solve for the unknown?
Ra = x Ta = z Rb = y Tb = w Then xz = 360 yw = 60 z = 3y w + z = 12 combining #1,3 ==> 3xy = 360 ==> xy = 120 Combining this with #2, xy = 120 and wy = 60 ==> x = 2w Combining #3,4 ==> z = 12 - w Substituting into #1, we get 2w( 12 - w) = 360 24w - 2w^2 = 360 0 = 2w^2 - 24w +360 0=w^2 -12w + 180 Using the quadratic formula, w = 6 + 12i or 6 - 12i x = 2w ==> x = 12 + 24i or 12- 24i z = 12 - w = 6 - 12i or 6 + 12i y = z/3 = 2 - 4i or 2 + 4i Ra =12 + 24i or 12- 24i Rb = 2 - 4i or 2 + 4i Ta = 6 - 12i or 6 + 12i Tb = 6 + 12i or 6 - 12i Using the first answer Ra -Rb + Ta - Tb = 10 + 4i I'll leave the evaluation of the second set of answers for you. I checked the first set of answers on the original equations, but left the second set for you todo.
Physics gravitational word problem?
the easiest way to do this is to use keplers laws. according to his third law, the square of the period of orbit over the cube of the radius of the orbit is constant for all objects orbiting the same body. so Ta^2/Ra^3=Tb^2/Rb^3 now since we know that mars' average radius is 1.52 times the radius of earth's, we can replace Rb^3 with (1.52Ra)^3 so: Ta^/Ra^3=Tb^2/(1.52Ra)^3 =Tb^3/3.511808Ra^3 so since Ra^3 appears on the bottom of both equations, we can cancel them both out. Ta^2=Tb^2/3.511808 Tb=SQRT(Ta^2*3.511808) =SQRT(365^2*3.511808) = about 684 days the accepted value is about 686, so thats not too bad edit: yes jim is mathematically correct, but the formula to begin with is not correct.
Please....someone help me to these math problem?
Runner A took 3 minutes and 45 seconds to complete a race and another runner B required 4 minutes to run the same race. The rate of the faster runner A is 0.4 meters per second more than the rate of the slower runner B. Find the rate of the faster runner A.
Why does the distance from the Sun vary the planets motion?
For the laws of Kepler.From the second law of Kepler2) A planet in rotating around the sun will sweep equal areas in equal timesit means that near the Sun (perihelion) the body will go faster while the aphelion will go less fast3) Kepler's lawR ^ 3 / T ^ 2 = constant = KOnly with Newton was it found that K = 4 π ^ 2 / GM but be careful that the value of K I wrote now does not belong to Kepler's law at that time K did not knowwe only knew that it was constant ... that is, the same for all the planets!Obviously M = mass of the Sun.G = 6.67 * 10 ^ -11 = universal gravity constantR = average planet distance -soleT = period of revolution of the planet around the sunThis Law (perhaps the most important!) Allows, known the constant K (which at the time of Kepler did not know but now we know thanks to Newton) to calculate R (and therefore know the size of the solar system if I know all the distances R from the sun) only verifying its period of revolution around the sun. Reclaimed T from observation and known K revenue RRA ^ 3 / RB ^ 3 = TA ^ 2 / TB ^ 2 from whichRa ^ 3 / Ta ^ 2 = Rb ^ 3 / Tb ^ 2if I take the earth as a reference point eRb = 1 AU and Tb = 1 terrestrial year thenRa ^ 3 / Ta ^ 2 = 1Ra ^ 3 = Ta ^ 2R = (T ^ 2) ^ 1/3in practice if you know the period of revolution of a planet around the Sun (just follow the motion of the planet in the sky for a certain time then know its period and immediately at what distance it is from the sun (and difference the distance from US )But R will be in UA (astronomical unit = 149 million Km) and T in Earth yearsIt should be specified that Kepler's laws are precise to the extent that the following hypotheses are met:1) the mass of the planet is negligible compared to that of the Sun;2) interactions between different planets can be ignored (such interactions lead to slight perturbations on the shape of the orbitsNote: English is not my first language, so I am really sorry for grammar errors and expression, so please feel free to use the "suggest edit" to help correct me. Thanks!
Two planets, A and B, are in a circular orbit around a star.?
We have g = GM/r^2 and g' = GM/R^2 that are offset by a = W^2 r and A = w^2 R so we can write: M is the common gravity source for both planets. g/g' = (M/M)(R/r)^2 = (W/w)^2 (r/R) = a/A (R/r)^3 = (1/t//1/T)^2 = (T/t)^2 R/r = (T/t)^(2/3) = (14)^(2/3) = 5.81 ANS.
How can I learn/memorise the atomic numbers of different elements?
Mnemonics for periodic table (groups)-LiNa Ki Ruby Se Friendship (Li,Na,K,Rb,Cs,Fr)Beta mange car scooter baap razi (Be,Mg,Ca,Sc,Ba,Ra)Baba Ali gaye india thailand (B,Al,Ga,In,Tl)Chemistry sir gives sunky problems (C,Si,Ge,Sn,Pb)Nana patekar, Aishwarya sab bimar (N,P,As,Sb,Bi)us side se teepo (O,S,Se,Te,Po)Faltu class, boring instructor (F, Cl, Br, I)Hena Nina aur Kareena, xerox rangeen. (He,Ne, Ar, Kr, Xe, Rn)Lanthanoides - sar par nadiya prem ki samayi, yuu gad-gad tab dil hua, are tum ye bhi lutoge? (Ce,pa,Nd,Pm,Sm,Eu,Gd,Tb,Dy,Ho,Er,Tm,Yb,Lu)Actinides Tume PadhaUUU newspaper, purane aam kam bike, cafe me aish farmate madonna larra (Th,Pa,U,Np,Pu,Am,Cm,Bk,Cf,Es,Fm,Md,No,La)Hope this helps. :)If you need mnemonics for biology. What are some mnemonics for NEET biology?