Av citi auon

What is the lift (in newtons) due to Bernoulli's principle on a wing of area 96 m2m2 if the air passes over the top and bottom s

egoroff_w [7]

Answer:

$F_{lift}=2.42*10^{9} N$

Explanation:

Apply Bernoulli's principle

$p_{b}+1/2pv_{b}^{2} =p_{t}+1/2pv_{t}^{2} \\p_{b}-p_{t}= 1/2pv_{t}^{2}-1/2pv_{b}^{2}\\p_{b}-p_{t}= 1/2(1.00*10^{3} kgm^{-3} )(270m/s^{2} )-1/2(1.00*10^{3}kgm^{-3} )(150m/s^{2} )\\p_{b}-p_{t}=2.52*10^{7}N/m^{2}\\ F_{lift}=2.52*10^{7}N/m^{2}*96m^{2} \\ F_{lift}=2.42*10^{9} N$

5 0

4 years ago

Suppose a distant world with surface gravity of 5.20 m/s2 has an atmospheric pressure of 7.16 ✕ 104 Pa at the surface. (a) What

Nesterboy [21]

Answer:

899752.13598 N

Explanation:

p = Pressure on the plate = $7.16\times 10^4\ Pa$

A = Area = $\pi r^2=\pi 2^2=\pi 4$

F = Force on the region

When force is divided by area we get pressure

$p=\frac{F}{A}\\\Rightarrow F=p\times A\\\Rightarrow F=7.16\times 10^4\times \pi 4\\\Rightarrow F=899752.13598\ N$

Force is exerted by the atmosphere on a disk-shaped region is 899752.13598 N

6 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀