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:
the object is decelerating
Answer:
6787.5 V
Explanation:
From the question,
P = IV..................... Equation 1
Where P = Power, I = rms current, V = rms voltage.
make V the subject of the equation
V = P/I................. Equation 2
Given: P = 1500 W, I = 6.4/√2 = 4.525 A
Substitute these values into equation 2
V = 1500(4.525)
V = 6787.5 V
Hence the rms voltage = 6787.5 V
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Answer:
20 ms¯¹
Explanation:
3. Determination of the final velocity
From the question given above, the following data were obtained:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
Acceleration is simply defined as the change in velocity per unit time.
Mathematically, it can be expressed as:
Acceleration (a) = final velocity – Initial velocity / time
a = v – u / t
With the above formula, we can obtain the final velocity of the car as follow:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
a = v – u / t
5 = v – 0 / 4
5 = v / 4
Cross multiply
v = 5 × 4
v = 20 ms¯¹
Thus, the final velocity of the car is 20 ms¯¹
