Answer:
0.75 m³/s
Explanation:
Applying,
Q = vA.................... Equation 1
Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.
From the question,
Given: v = 2.5 m/s, A = 0.3 m²
Substitute these values into equation 1
Q = 2.5(0.3)
Q = 0.75 m³/s
Hence the flow rate of water in the pipe is 0.75 m³/s
Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.
The value of the thrust developed by the engine of a boeing 777 in N and Kgf are ;
i) 376616N
ii) 38430Kgf
<h3>What is force?</h3>
Force is a push or a pull. The reactive force always serve to balance the applied force. We are here asked to convert the the thrust developed by the engine of a boeing 777 which is about 85400 lbf to the following units;
i) N
ii)kgf
Thus;
1 Ib = 0.45 Kg
1 lbf = 0.45 Kg * 9.8 m/s^2 = 4.41 N
We know that;
1 lbf = 4.41 N
85400 lbf = 85400 lbf. * 4.41 N/1 lbf
= 376616N
Again;
1 lbf = 0.45 Kgf
85400 lbf = 85400 lbf * 0.45 Kgf/1 lbf
= 38430Kgf
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Answer:
110 m
Explanation:
Draw a free body diagram of the car. The car has three forces acting on it: normal force up, weight down, and friction to the left.
Sum of the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of the forces in the x direction:
∑F = ma
-F = ma
-Nμ = ma
Substitute:
-mgμ = ma
-gμ = a
Given μ = 0.40:
a = -(9.8 m/s²) (0.40)
a = -3.92 m/s²
Given that v₀ = 30 m/s and v = 0 m/s:
v² = v₀² + 2aΔx
(0 m/s)² = (30 m/s)² + 2 (-3.9s m/s²) Δx
Δx ≈ 110 m
