Answer:
a) 60 N
b) 860 N
Explanation:
Given that,
= 100 kg
= 20 kg
= 8.0
= 3.0
a) By Newton's Law,
∑
∑
Hence,
b) By Newton's Law
∑
Hence,
Net force acting on 100 kg mass,
rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind t
unnel. The freestream velocity is 40 m/s. The density of the air is 1.2250 kg/m3, and the absolute viscosity is 1.7894 x 10-5 kg/m s. Plot (not sketch) the velocity profiles at x = 0.0 m, x = 0.05 m, x = 0.10 m, and x = 0.20 m. Calculate the chordwise distribution of the skin friction coefficient and the displacement thickness. What is the drag coefficient for the plate? Be sure to account for both sides of the plate. Ans.: CD = 3.587 x 10
1 answer:
Answer:
See explaination for the details of the answer.
Explanation:
Density is a measurement that compares the amount of matter an object has to its volume. An object with much matter in a certain volume has high density An object with little matter in the same amount of volume has a low density. Density is found by dividing the mass of an object by its volume.
The Calculate of the chordwise distribution of the skin friction coefficient and the displacement thickness is done and represented in the attachment.
Please kindly check attachment for the step by step answer.
You might be interested in
Answer:
n=0.03928 moles
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Explanation:
The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs
Volume of oxygen in lungs =V=5*20%= 1 L=
Temperature=T=
Pressure at sea level = P= 1 atm=
R is universal Gas Constant =8.314 J/mol.K
Formula:
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles