Answer- I would think low
The attraction will decrease. Hope I helped :)
Answer:
Explanation:
Given that,
Surface area A= 17m²
The speed at the top v" = 66m/s
Speed beneath is v' =40 m/s
The density of air p =1.29kg/m³
Weight of plane?
Assuming that,
the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.
Using Bernoulli equation
P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''
Where
P' is pressure at the bottom in N/m²
P" is pressure at the top in N/m²
v' is velocity at the bottom in m/s
v" is velocity at the top in m/s
Then, Bernoulli equation becomes
P'+ ½pv'² = P'' + ½pv''²
Rearranging
P' — P'' = ½pv"² —½pv'²
P'—P" = ½p ( v"² —v'²)
P'—P" = ½ × 1.29 × (66²-40²)
P'—P" = 1777.62 N/m²
Lift force can be found from
Pressure = force/Area
Force = ∆P ×A
Force = (P' —P")×A
Since we already have (P'—P")
Then, F=W = (P' —P")×A
W = 1777.62 × 17
W = 30,219.54 N
The weight of the plane is 30.22 KN
Answer:
Im actually depressed
Explanation:
jk thx for the points thoo
Answer:
Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...
Explanation:
