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
The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls
Answer:
it snaps
Explanation:
the more force you put on it, the wired out it gets than it snaps. I think
Kinetic Energy = (1/2) (mass) (speed)
First runner: KE = (1/2) (45kg) (49 m/s) = 1,102.5 Joules
Second runner: KE = (1/2) (93kg) (9 m/s) = 418.5 Joules
The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.
