The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Explanation:
Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,
...........(1)
Given, h = 4, x = 40 and s(t) = -20 mph
Differentiate equation (1) wrt t


When x = 40, 



So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.
<span>1 C = 6.24150965(16)×10^18 electrons
31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs
</span><span>I hope this helps. </span>
<span>4) Formation of a gas
When methane burns, it produces the gases water vapor and carbon dioxide.
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