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Kosy
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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
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Two cars x and y travelling side by side with a common velocity are subjected to constant retardation of 1/2m/s^2 and 1/6m/s^2
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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
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Answers (27)
the image is at infinity so d=fo+fe=100+3=103cm
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 acceleration
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
time for rock to fall plus time for sound to travel to him equals 2.8s.solve both.add them and solve for H
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
Deceleration=0.74*10=7.4m/s^2 v-u=at v-19.5=-7.4 v=-7.4+19.5=12.1m/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
v^2=rg v=sqrt(42000*1*10^-3)=6.5m/s
since it without replacement total no of ball reduces p(OY)=(4/13*6/12)+(6/13*4/12)=12/33
F(x)=f(2x) x^2=f(2x) f(2x)=(2x)^2=4x^2
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
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
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
Formula:E=mc^2
Dk=33sin39+1.5 Dk=20.8+1.5=22.3m
sorry, I mean m/s^2
People who attended=40-16=24 people.percentage=24/40*100%=60%
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
H=u^2/2g H=60^2/20=3600/20=180m
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
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
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
E=mc*temp change 50,000=M*4200*(75-17) M=50,000/4200*58 M=0.205Kg
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
200+1200/200=700m/s
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)
First of all,calculate acceleration.then use the acceleration the net force.then subtract the net force from F1.
