0.289352 days is the 25000 seconds.
<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let
be the pressure at a point.
Let
be the density fluid at a point.
Let
be the velocity of fluid at a point.
Bernoulli's equation states that
for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let
be the pressure of a point just above the wing.
Let
be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.
![\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%281.29%29%28255%29%5E%7B2%7D%2Bp_%7Bup%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%281.29%29%28199%29%5E%7B2%7D%2Bp_%7Bdo%7D)
So,![p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa](https://tex.z-dn.net/?f=p_%7Bup%7D-p_%7Bdo%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201.29%5Ctimes%20%2825424%29%3D16398.48Pa)
Force is given by the product of pressure difference and area.
Given that area is
.
So,lifting force is ![16398.48\times 27=442758.96N](https://tex.z-dn.net/?f=16398.48%5Ctimes%2027%3D442758.96N)
Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
Answer:
-26 m/s.
Explanation:
Hello,
In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:
![t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B0m%2Fs-26m%2Fs%7D%7B-9.8m%2Fs%5E2%7D%20%3D2.65s)
With which we compute the maximum height:
![y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m](https://tex.z-dn.net/?f=y%3D26m%2Fs%2A2.65s-%5Cfrac%7B1%7D%7B2%7D%2A9.8m%2Fs%5E2%2A%282.65s%29%5E2%20%5C%5C%5C%5Cy%3D34.5m)
Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:
![v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s](https://tex.z-dn.net/?f=v_f%3D%5Csqrt%7B0m%2Fs-%28-9.8m%2Fs%5E2%29%2A2%2A34.5m%7D%5C%5C%20%5C%5Cv_f%3D-26m%2Fs)
Which is clearly negative since it the projectile is moving downwards the starting point.
Regards.
He proposed the sun-centered model of the solar system.