Thakgalo sakwana

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

Questions

The following questions were asked by visitors named Thakgalo sakwana.

An object is projected vertically upwards and returns to the point of projection 12s later. Ignore air resistance and determine the velocity with which it was projected?

Answers

The following answers were posted by visitors named Thakgalo sakwana.

Vf=Vi +at Vf=0, Vi=?, a=-9.8 remeber time taken to go up is equal to the time taken to go back to the point of projectile.. THEREFORE t=6 Vf=Vi +at (0)=Vi +(-9.8)(6) Vi=58.8m/s ***[OR]*** Vf=Vi +at Vf=-Vi, Vi=?, a=-9.8, t=12 (-Vi)=Vi +(-9.8)(12) -2Vi= -11...

Sin¤= 1/3 ¤=19.47' Fnet= Fg// - Ff Fnet= mg.sin¤ -Ff =2(9.8)sin19.47-0.35 Fnet= 6.183N a= Fnet/m a= 6.183/2 a=3.091m/s^2 (Vf^2)=(Vi^2) +2as Vf^2= (0)+2(3.091)(3) Vf^2= 18.546 Vf= 4.31m/s downward. ***GOOD LUCK*** For info call me on 0723624826

100KHP/(KHP+NaHO) =100(0.847)/(0.847+19.82) =4.098...%

Vf= vi +at vf= (0) +(-10)(10) vf=-100m/s downward