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
They send out waves differently and cannot be heard easily
We know that whoever she is is traveling to Antarctica or elsewhere
in the south polar region. June is the beginning of Winter there, with
zero to extremely short daylight.
But we still don't know her name.
Answer:
Mass = 18.0 kg
Explanation:
From Hooke's law,
F = ke
where: F is the force, k is the spring constant and e is the extension.
But, F = mg
So that,
mg = ke
On the Earth, let the gravitational force be 10 m/.
3.0 x 10 = k x 5.0
30 = 5k
⇒ k = ................ 1
On the Moon, the gravitational force is of that on the Earth.
m x = k x 5.0
= 5k
⇒ k = ............. 2
Equating 1 and 2, we have;
=
m =
= 18.0
m = 18.0 kg
The mass required to produce the same extension on the Moon is 18 kg.