Aflat plate is oriented parallel to a 45 m/s airflow at 20°c and atmospheric pressure. the plate is l = 1 m in the flow direction and 0.5 m wide. on one side of the plate, the boundary layer is tripped at the leading edge, and on the other side there is no tripping device. find the total drag force on the plate.

4.192 N

Explanation:

Step 1: Identify the given parameters

Velocity of airflow = 45m/s

air temperature = 20⁰

plate length and width = 1m and 0.5m respectively.

Step 2: calculate drag force due to shear stress,

Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²

⇒The Reynolds number () based on the length of the plate is

= 3 X10⁶  (flow is turbulent, Re ≥ 500,000)

⇒The average shear stress coefficient () on the "tripped" side of the plate is

= 0.0038

⇒The average shear stress coefficient () on the "untripped" side of the plate is

= 0.0031

The total drag force =

The total drag force is 4.192 N

gets less

  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