Answer:
the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.
w is angular frequency
formula:2π/T
now k is spring constant
F/R-->mw²
putting values:70*(2π/0.2)²
=4.9x10⁶
so we can say that SHM is not affected by the amplitude of the bounce.
Given:
u = 6.5 m/s, initial velocity
a = 1.5 m/s², acceleration
s = 100.0 m, displacement
Let v = the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s
Answer: 18.5 m/s
a tree truck
is the answer i hope this helps you xD
Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
![(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]](https://tex.z-dn.net/?f=%280.5%29%5E%7B2%7D-%2810%29%5E%7B2%7D%20%3D%202%2Aa%2A%28100%29%5C%5C0.25-10000%3D200%2Aa%5C%5C200%2Aa%3D-9999.75%5C%5Ca%20%3D-50%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign of acceleration means that the ship slows down its velocity in order to land.