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Answer:
it is safe to stand at the end of the table
Explanation:
For this exercise we use the rotational equilibrium condition
Στ = 0
W x₁ - w x₂ - w_table x₃ = 0
M x₁ - m x₂ - m_table x₃ = 0
where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m
the mass of the man is 62 kg and the distance
x₂ = 4.5 - 0.85
x₂ = 3.65 m
the mass of the table (m_table = 22 kg) is at its geometric center
x_{cm} = L/2 = 2.25 m
x₃ = 2.25 -0.85
x₃ = 1.4 m
let's look for the maximum mass of man
m_{maximum} =
let's calculate
m_{maximum} = (380 0.85 - 22 1.4) / 3.65
m_{maximum} = 80 kg
we can see that the maximum mass that the board supports without turning is greater than the mass of man
m_{maximum}> m
consequently it is safe to stand at the end of the table
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>
Answer:
F = 0
Explanation:
Given that,
Force acting to push a bin = 45 N
Mass of the plastic bin, m = 3 kg
The plastic bin is moving with a constant velocity.
We need to find the net force acting on the box. Constant velocity means the change in velocity is equal to 0. It means acceleration will be 0.
As a result, the force acting on the box is equal to 0.