Amassless rod of length l is hinged with a frictionless hinge at one end. at the end opposite to the hinge there is a very small mass m attached to the rod. the rod is released from rest in the horizontal position. what is the frequency f when the rod passes through the vertical position? 1. 2πsqrt(g/l)2.2g/l3.(1/2π)sqrt(2g/l) 4. (1/2π)sqrt(g/l)5. 2πsqrt(2g/l)6. sqrt(2g/l)7.g/l8. sqrt(g/l)

4.

Explanation:

We have in this case a simple pendulum, of length L and mass m. As we know, the period of this pendulum is:

but it is true only in small oscillation angles, it meas less than 1 rad.

The frequency is the inverse of the period, then:

The correct answer is 4.

I hope it helps you!

  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

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