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
Answer:F(of gravity) = MA
F(normal force) = MA * cos(angle)
F = 72 * 9.81 * cos28
Don't have a calculator, so can't really do all the math right there. So just plug that in
Explanation:
i dont really know
For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:

where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:

, so the work done by the brakes to stop the bus is

And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
The first one, as the mass is higher so it accelerates more