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II Newton's law: a=F/m a=16N/4kg a=4m/s^2 a=Δv/Δt Δv=a x Δt Δv=4m/s^2 x 2s Δv=8m/s change in velocity is 8m/s
d/dt (2xy-4x/3y^3)= d/dt (48) 2*d(xy)/dt - d(4x/3y^3)/dt = 0 (because derrvation of constant is 0) 2* ((dx/dt)*y + xdy/dt) -(((d(4x)/dt)*3y^3^-4x*d(3y^3)/dt))/(3y^3)^2)=0 2*((-6)*(-2)+3*(dy/dt))-(4*(dx/dt)*(3y^3)-4*x*3*(3*y^2)*(dy/dt) )/9y^6 =0
x-number the adult tickets y-number the child tickets x+y=550 5.50*x + 3.50*y =2139.00 ---------- y=550-x 5.50*x +3.50*(550-x) =2139.00 ----- 5.50*x +3.5*550- 3.50*x =2139.00 (5.5-3.5)*x +1925.00 = 2139.00 2*x=2139.00-1925.00 2*x=214 x=214/2 x=107 107
No second ball has the same acceleration (acceleration of Earths gravity) but it (ball) has initial velocity. It is calculable knowing that in moment of overtaking both balls had the same traveled distance. g*t^2/2= V*t + g(t-2)^2/2 (t-2) is because the

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