A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?

The value is  

Explanation:

From the question we are told that

    The  acceleration is

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

Here   is the frictional force which is mathematically represented as

substituting into above equation

=>        

substituting values

the answer is b

explanation:

  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