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:

Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
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Where is the rest .........
Gravitational attraction / field strength increases when closer; A light dependent resistor (LDR) can be used as a sensor to detect light intensity.