Im confused bc im not sure if this is a full question
At stp conditions (
), the speed of sound is
The sound wave moves by uniform motion, so we can use the basic relationship between space, time and velocity:
where S is the distance covered by the sound wave in a time t. In our problem, t=3.00 s, therefore the distance covered by the sound wave is
Answer:
120N
Explanation:
Newton's second law formula: F= ma
given that m = 10 kg, a = 12 m/s^2
F = ma = 10 kg * 12 m/s^2 = 120 kgm/s^2 = 120 N
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
It would be A. Because think of the explanations Jasons friend could say to them that would be a negative 'statement'.