An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t
he kinetic energy becomes KE2. How can you relate KE2, to KE1?
Answer and I will give you brainiliest
Answer and I will give you brainiliest
1 answer:
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Explanation:
Given that,
2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.
The tension in one of them is then decreased sightly, then will decrese.
Beat frequency,
So, the new frequency of the string is 217 Hz. Hence, this is the required solution.
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 =
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