Answer:
Initial pressure = 6 atm. Work = 0.144 J
Explanation:
You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J
Answer:
First one, third one, and fourth one
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:
The boundary layer thickness is equal to:
ft
The shear stress is equal to:
b) If the railing edge is 2 ft, the Reynold number is:
The boundary layer is equal to:
The sear stress is equal to:
c) The drag coefficient is equal to:
The friction drag is equal to:
Life Expectancy Was Shorter
Trains were faster to ride
The population explosion was not as on date
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation =
sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
