If you hold a 3.89-kg package by a light vertical string, what will be the tension in this string when the elevator accelerates as in the previous part?

a) 784.5 N

b) 47.8 N

Explanation:

a)

Find the free-body diagram of the person in attachment.

There are two forces acting on the person:

- Its weight (gravitational force), of magnitude W = 625 N, downward

- The normal reaction exerted by the scale on the person, upward, of magnitude N: this corresponds to the reading on the scale

Therefore, the net upward force on the person is

According to Newton's second law of motion, the net force must be equal to the product of mass and acceleration, so:

where:

is the mass of the person

is the acceleration of the elevator (upward)

Re-arranging the equation and solving for N, we can find the reading on the scale (the normal force):

b)

In this case, you are holding a package of 3.85 kg with a light string.

The forces acting on the package are:

- The weight (force of gravity), , downward

- The tension in the string, T, upward

As before, the net force on the package is

And according to Newton's second law, it must be equal to the product of mass and acceleration:

Therefore here we have:

m = 3.89 kg is the mass

is the acceleration due to gravity

is the acceleration of the elevator

Solving for T, we find the tension in the string:

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  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