Electrostatic forces are non-contact forces; they pull or push on objects without touching them
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = ![\frac{\Delta F}{A \ \rho \ g} + v_1^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20F%7D%7BA%20%5C%20%5Crho%20%20%5C%20g%7D%20%2B%20v_1%5E2)
suppose that the area of the wing is A = 1 m²
we substitute
v₂² =
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s
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Hope this helps. <3
The weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.
<h3>
Principle of moment</h3>
The weight of the second object is determined by applying principle of moment as shown below;
take moment about the pivot,
sum of clockwise moment = sum of anticlockwise moment
F₁r₁ = F₂r₂
5 x 0.5 = F₂ x 1
2.5 N = F₂
Thus, the weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.
Learn more about moment here: brainly.com/question/6278006