Answer:
B). 3.4 s
Explanation:
As we can see the graph is given between velocity and time
so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed
Then it's speed decreases to next few second and then speed increases to its maximum value
The time after which velocity comes to its maximum value will reach after t = 3 s
so out of the all given options most correct option will be
This question is incomplete, the complete question is;
Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.
Answer:
- the magnitude of the resultant force is 557.32 kN
- the direction of resultant force is 48.29°
Explanation:
Given the data in the question and the diagram below,
First we work on the force on the left hand side.
Left Horizontal
= βgAr
here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²
we substitute
= βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N
Left Vertical
= ( βgπh² / 2 ) × W
we substitute
= [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N
Now we go to the right hand side
Right Horizontal
= βgAh
here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²
we substitute
= ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N
Right Vertical
= ( βgπh² / 4 ) × W
we substitute
= [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N
Hence
Fx = -
= 52974 N - 423792 N = 370818 N
Fy = +
= 332845.458 N + 83211.36 N = 416056.818 N
R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N
R = 557.32 kN
Therefore, the magnitude of the resultant force is 557.32 kN
Direction of resultant force;
tanθ = Fy / Fx
we substitute
tanθ = 416056.818 N / 370818 N
tanθ = 1.121997
θ = tan⁻¹( 1.121997 )
θ = 48.29°
Therefore, the direction of resultant force is 48.29°
