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jaykim
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From triangle,cotu=x/5 differentiate both side cosec^2udu/dt=dx/dt/5subtitute u and du/dt and solve for dx/dt
H=ho+ut-1/2gt^2 when H=0 t=4secs 0=35.3+4u-80 4u=44.7 u=11.2m/s
for x-component:0.04*2.55sin45=0.04v1sin22.5+0.04v2sin35.9 for y-component:0.04*2.55cos45=0.04v1cos22.5+0.04v2cos35.9 1.803=0.38v1+0.59v2 1.803=0.92v1+0.81v2 solve for v1 and v2
1)10/26=X/150 X=1500/26=57.7m 2)costheta=10/26 theta=67.3degrees
Q=mcdt=500/1000*400*(120-30)=18000J
the velocity with which it hit the ground vo=sqrt(2*10*5)=10m/s hn=e^nvo^2/2g 2=e^3(100)/2(10) e^3=0.4 e=0.632
H=10t-5t^2 10-H=10t+5t^2 10-10t+5t^2=10t+5t^2 20t=10 t=0.5secs H=10(0.5)-5(0.25)=3.75m
F=13.0m1 F=3.6m2 therefore 3.6m2=13m1 m2=3.61m1 F=(3.61m1-m1)a F=2.61m1a m1=F/13 F=2.61*(F/13)a 2.61a=13 a=4.98m/s^2
wi=181rpm=2pi(181/60)=18.95rad/s wf=260rpm=2pi(260/60)=27.23rad/s angular acceleration=a/r=wf-wi/t a/0.285=27.23-18.95/19.9 3.51a=0.416 a=0.12m/s^2
Cofficient of static friction=tan28=0.53 (b)mgsin28-ukmgcos28=ma cancel m d=1/2at^2 2.58=1/2a(3.92)^2 a=0.336m/s^2 substitute a:4.695-8.829uk=0.336 8.829uk=4.359 uk=0.49
Let u=2x du/dx=2 =1/2Incos^7u(int=integral) 1/2Intcos^6u*cosu since cos^2u=1-sin^2u there cos^6u=(cos^2u)^3=(1-sin^2u)^3 there the integral become:1/2Int(1-sin^2u)^3cosudu let v=sinu dv/du=cosu du=dv/cosu substitute it in du and v in integral
v=dx/dt=2t^2 dx=2t^2dt integrating both side gives: x=2/3t^3+c substituting t=0 and x=2.8 gives c=2.8 therefore x=2/3t^3+2.8 so when t=2.7 x=2/3(2.7)^3+2.8=15.9m
fo/fe=3.11 fe=1.65/3.11=0.53m since at infinity,their focal length coincides,d=1.65+0.53=2.18m
a=0(starting from a corner pole1) no of poles=n=72 t7=42(corner pole 2) d=spacing 42=0+d(7-1) 6d=42 d=7m
Fx=67.3+98.7cos45+133cos315=231.1N Fy=0+98.7sin45+133sin315=-24.25N R=sqrt(231.1^2+24.25^2)=232.4N
Let G=integral cosxsinxsqrt(1+sinx)dx let u=1+sinx du/dx=cosx dx=du/cosx substituting dx cancel cos x in the integral so It become:integral(u-1)sqrtudu(since sinx=u-1) therefore open bracket gives:u^1/2(u-1)du=u^3/2-u^1/2du integrate each
Formula is:d/po=nL/nw d=31.1-27.6=3.5 po=31.1 nw=390/18=21.7moles 3.5/31.1=nL/21.7 nL=2.442moles=86.7/M M=35.5g/mol
there is a tension in string its vertical component:Tsintheta=mg while Tcostheta=mv^2/r dividing equ1 by equ2 gives:tantheta=gr/v^2 tan35=10*0.6/v^2 v^2=8.57 v=2.93m/s
W=165*30*0.8=3960J
Differentiate y=ax^3 gives:dy/dx=3ax^2 from 3x+y=b rearranging y=-3x+b m=-3 therefore 3ax^2=-3 ax^2=-1 since x=-3 a(-3)^2=-1 9a=-1 a=-1/9 (b)when x=-3 y=0 0=-3(-3)+b b=-9
Let u=7x du/dx=7 du/7=dx therefore,1/7integral sinucosudu let v=sinu dv/du=cosu du=dv/cosu: 1/7intergralvdv=1/7v^2/2=1/14sin^2(7x)
component of paul force:31cos45i+31sin45j=21.9i+21.9j component of johny force:52cos216i+52sin216j=-42.1i-30.6j Connie force is the resultant=21.9i+21.9j-42.1i-30.6j=-20.2i-8.7j direction=arctan(-8.7/-20.2)=23.3degree Fmag=sqrt(20.2^2+8.7^2)=21.99N
In 2 hours without stopping,bus A travels 90*2=180km since bus B stopped for a total of 20 minutes,it is clear it only travelled for 2-1/3=1hour 40mins distance it travelled =5/3*60=100km distance at ten a.m=360-180-100=80km
x+y=243 1/2x=4y x=8y 8y+y=243 9y=243 y=27 x+27=243 x=216
1)v=sqrt(2*9.8*3.14sin40)=6.29m/s (2)umgd=1/2mv^2 u*9.8*4.68=0.5(6.29)^2 u=0.43 (3)change in mechanical energy=workdone by friction=0.43*11.7*9.8*46.8=2307.4J
if benzoic acid ionize partially to x mole of both H30+ and C6H5COO-.At equilibrium, we have 0.1-xM of benzoic acid:6.4*10^-5=x^2/0.1-x since x is relatively small we have:x^2=6.4*10^-5*0.1 x=2.5*10^-3M
volume of ice cube is 3^3=27cm^3 density of ice=0.9167g/cm^3 mass of ice=27*0.9167=24.8g=0.0248kg heat energy=mL=0.0248*336000=8333J
nos are n,n+2andn+4 3(n+2)=7+n+n+4 3n+6=11+2n n=5 so the nos are 5,7 and 9
distance of each charge to the centre of rectangle,d=sqrt(3.5^2+2^2)=4.031m A charge contribute in the potential V=9*10^9*2.51*10^-9/4.031 V=5.604v since d charge are equal and are at equal distance from centre,the electric potential at
using doppler equation:f2=f1[Vw/(Vw-Vs)] where f2=450hz f1=360hz Vw=340m/s Vs=? 450=360(340/340-vs) 1.25(340-vs)=340 340-vs=272 vs=68m/s
Let assume mass of glucose consumed as Mg converting to moles give:M/180.2moles.so the heat energy released by oxidation of this amount of glucose is:M/180.2*2.82*10^6JThis heat cause the temp rise of 3celsius. 15649.2M=80*4180*3 M=64.1Kg
w=ar*t 18=4ar ar=4.5rad/s^2 (b)18^2=2ar*theta theta=36rad
