30) a foul ball is hit straight up into the air with a speed of 30.0 m/s. (a) calculate the time required for the ball to rise to its maximum height. (b) calculate the maximum height reached by the ball. (c) determine the time at which the ball pass a point 25.0 m above the point of contact between the bat and ball. (d) explain why t

a) 3.05 seconds

b) 45.87 m

c) 0.995 seconds

Explanation:

t = Time taken

u = Initial velocity = 30 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = 9.81 m/s² (positive downward, negative upward)

a) Time taken by the ball to reach maximum height

Time taken by the ball to reach maximum height is 3.05 seconds

b) Maximum height

Maximum height reached by the ball is 45.87 m

c) If s = 25 m

Velocity of the ball at 25 m height is 20.24 m/s

Time taken by the ball to reach a point 25.0 m above the point of contact between the bat and ball is 0.995 seconds

  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

true is the correct answer

hope i