What is the average speed, over the first 4.0 s of its motion, of a pebble released from rest off a bridge?

 Average speed, over the first 4.0 s of its motion, of a pebble released from rest off a bridge = 19.6 m/s

Explanation:

We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case the initial velocity of body = 0 m/s, acceleration = 9.8 , we need to calculate displacement when t = 4 seconds.

Substituting

So a distance of 78.4 meter has been traveled by pebble in 4 seconds.

Average speed = 78.4/4 = 19.6 m/s

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

c. magnitude and direction