A merry-go-round rotates from rest with an angular acceleration of 1.30 rad/s2. (a) How long does it take to rotate through the first 3.00 rev? s (b) How long does it take to rotate through the next 3.00 rev? s

(a) t=5.384s

(b) t=2.23s

Explanation:

Given data

Initial angular speed of merry go round ω₀=0 rad/s

Angular acceleration of merry go round α=1.30 rad/s²

For Part (a)

Using the equation  of angular motion for constant angular acceleration

Θ=ω₀t+(1/2)at²

The time that the merry takes to complete 3 revolutions is 5.384 seconds

For Part (b)

Using the equations of angular motion for constant acceleration to find the time for the round to complete 6 revolutions

Θ=ω₀t+(1/2)at²

The time that the round takes to finish the second three revolutions is

  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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)