To solve this problem, we are going to use the formula for
work which is Fd where x and y are measured separately.
X direction: W = 13.5 x 230 = 3105 Joules
Y direction: W = -14.3 x -165 = 2360 Joules
So the total work is getting the sum of the two: 3105 + 2360
= 5465 Joules
Answer: 0.0163
Explanation: fn=n x v/2L
Fundamental frequency =f1
1 x (343/2L) = 10500
Rearrange the equation
L= v/(2xFn)
L= 343/(2x10500)
L=0.0163
Answer:
<em>1108.464 N of force</em>
Explanation:
diameter of water hose = 70 cm = 0.7 m
radius = 0.7/2 = 0.35 m
volumetric flow rate Q = 420 L/min
1 L = 0.001 m^3
1 min = 60 s
therefore,
Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s
Area A of fire hose = π
= 3.142 x
= 0.38 m^2
<em>From continuity equation, Q = AV</em>
where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.
Q = A1V1
0.007 = 0.38V1
V1 = 0.007/0.38 = 0.018 m/s.
Nozzle diameter = 0.75 cm = 0.0075 m
radius = 0.00375
Area = π
= 3.142 x
= 4.42 x
m^2
velocity of water through the nozzle will be
V2 = Q/A2 = 0.007 ÷ (4.42 x
) = 158.37 m/s
From
<em>F = ρQ(v2 - v1)</em>
Where,
F = force exerted
p = density of water = 1000 kg/m^3
F = 1000 x 0.007 x (158.37 - 0.018) = <em>1108.464 N of force</em>
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