Frequency and velocity of a particle in simple harmonic motion problem in that the amplitude is 0.24 in. and the maximum acceleration is 225 ft/s maximum velocity of the particle ist/s, and the frequency of its motion s z print 20r

velocity maximum = 21.6 ft /s

frequency = 16.88 Hz

Explanation:

Amplitude, A = 0.24 in = 0.02 ft

maximum acceleration, a = 225 ft/s\

The formula for maximum acceleration is

a = ω² A

225 = ω² x 0.02

ω² = 11250

ω = 106.06 rad/s

Maximum velocity, v = ω A

v = 106.06 x 0.02 = 2.1 ft/s

Let f be the frequency

ω = 2 x 3.14 x f

f = 106.06 / (2 x 3.14) = 16.88 Hz

answer; i believe the correct answer is (40,000 j) good luck

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂