Determine the resultant horizontal and vertical force component that the water exerts on the side of the dam. The dam is 25 ft l
ong and specific weight = 62.4 ib/ft cubed. Diagram is located at:_________
1 answer:
Answer:
Your question is not complete. However, i have provided the missing part both the diagram and remaining data as stated in attached full.
Resultant vertical force component is 260005.2 Ibs
Resultant horizontal force component is 487500 Ibs
Explanation:
From P = Specific Weight * height
P = 62.4 * 25 = 1560Ib/ft^2
Then, W = P * Z
W = 1560 * 25 = 39000Ib/ft
Resultant forces
Let the base length be 10ft
A = 2/3 (a*b)
A = 2/3 (10* 25) = 166.67ft^2
Resolving vertical component
F vertical = Specific weight * Volume
F vertical = 62.4 (166.67 * 25)
F vertical = 260005.2 Ibs
Resolving horizontal component
F horizontal = 1/2 (W * height )
F horizontal = 1/2 ( 39000 *25)
F horizontal = 487500 Ibs
You might be interested in
Answer:
See it in the pic
Explanation:
See it in the pic
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
