Kosy

This page lists questions and answers that were posted by visitors named Kosy.

Questions

The following questions were asked by visitors named Kosy.

Two cars x and y travelling side by side are subjected to constant retardations 1/2m/s and 1/6m/s.After passing a point A,they come to rest 300m before and 150m beyond a second point B.Find common velocity at A and distance AB

XYZ are three points on a straight road.A car passes X with a velocity of 5m/s.It travels from X to Y with a constant acceleration of 2m/s^2 and from Y to Z with constant retardation of 3.5m/s^2.find the velocity of the car as it passes Y and find distanc...

Answers

The following answers were posted by visitors named Kosy.

First of all,calculate acceleration.then use the acceleration the net force.then subtract the net force from F1.

Centre(0,0) divide the equ by 9.frm xsqr/9+ysqr/1=1 asqr:9 a-3 bsqr:1 b-1 vertices:(3,0)(-3,0)(0,1)(0,-1) eccentricity:e-sqrt(1-1/9)=2sqrt2/3 foci=(ae,0)(ae,0)=(2sqrt2,0)(-2sqrt2,0)

200+1200/200=700m/s

Let numbers be x and y where y is greater y-x=7-equ(1) 60/x-60/y=7-equ(2) simplify equ 2:60y-60x/xy=7 60(y-x)/xy=7-equ 3 subt equ1 in equ3:420=7xy 420=7x(7+x) divide both sds by 7:60=x(7+x) x^2+7x-60=0 x(x+12)-5(x+12)=0 x=+5 y=7+5=12

E=mc*temp change 50,000=M*4200*(75-17) M=50,000/4200*58 M=0.205Kg

100I=PRT 10,000*6.3*3/100=1890 In 3 month she is supposed to pay(300*3)=900 interest paid=1890-900=990

Electrostatic force=centripetal force Fe=9*10^9*4*(1.6*10^-19)^2/(2.26*10^-11)^2=1.804*10^-6N 1.804*10^-6=9.1*10^-31a a=1.983*10^-24m/s^2

10 years ago H-10=2(D-10)equ 1 H=D+12 equ 2 sub equ 2 in equ 1 D+12-10=2D-20 D+2=2D-20 D=22yrs H=22+12=34yrs

H=u^2/2g H=60^2/20=3600/20=180m

mark does 1/40 of the job and manny does 1/30 of the job. Together:1/40+1/30=7/120 of the job in a minutes,time to do the entire job is 120/7=17mins approx

People who attended=40-16=24 people.percentage=24/40*100%=60%

sorry, I mean m/s^2

Dk=33sin39+1.5 Dk=20.8+1.5=22.3m

h/d=sin7 d=100/sin7=820.6m v^2=2gd(sin7-0.11cos7)=2*9.8*820.6*0.0127 v=sqrt(204.264) v=14.3m/s k.e at bottom=1/2*80*14.3^28179.6joules

force in x-direction Fx:5cos0+12cos90+7cos180+12cos360=10kN Fy:5sin0+12sin90+7sin180+12sin360=12kN Net force=sqrt(10^2+12^2)=15.6kN

v=rw w=2.42/0.19=12.74rad/s r=2.88/12.74=0.226m 2)Banking angle=v^2/rg=27.7^2/800g=61.1^2/rg r=(61.1/27.7)^2*800=3892m F=1620(61.1)(61.1)/3892=1554N

F(x)=f(2x) x^2=f(2x) f(2x)=(2x)^2=4x^2

since it without replacement total no of ball reduces p(OY)=(4/13*6/12)+(6/13*4/12)=12/33

v^2=rg v=sqrt(42000*1*10^-3)=6.5m/s

Since the body moves with constant velocity,Pushing force equals frictional force pushing force in x-direction=fcostheta=0.27mg costheta=2.7m/f when pushing force can't do work is when it equals sum of weight of body and frictional forcef=mg+0.27mg costhe...

Deceleration=0.74*10=7.4m/s^2 v-u=at v-19.5=-7.4 v=-7.4+19.5=12.1m/s

Differentiate the equation: 2x+2ydy/dx=0 2ydy/dx=-2x dy/dx=-x/y dy/dx is the gradient m substitute for x and y m=1/sqrt3 equation for tangent line is y-y1=m(x-x1) where x1 and y1 are -1 and sqrt3 substitute it in equation b)when dy/dx=0 x=0 c)when dy/dx e...

time for rock to fall plus time for sound to travel to him equals 2.8s.solve both.add them and solve for H

after 3hrs the distance travelled by A increases by 19*3=57 nautical miles.Distance by A equals 50+57=117nm distance by B equals 24*3=72nm.distance between them is sqrt(117^2+72^2) equals 137.4nm

parameter:m-137kg r-0.901m coeff-0.353m normal force-109N angular velocity-413rpm=43.25rad/s Torque exerted by friction=moment of inertiaI*angular acceleration Ffr=0.353*109=38.48N I=1/2MR^2=0.5*137*0.901^2=55.61Kgm^2 38.48*0.901=55.61*angular acceleratio...

the image is at infinity so d=fo+fe=100+3=103cm