Answer:
91.84 m/s²
Explanation:
velocity, v = 600 m/s
acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2
Let the radius of the loop is r.
he experiences a centripetal force.
centripetal acceleration,
a = v² / r
39.2 x r = 600 x 600
r = 3600 / 39.2
r = 91.84 m/s²
Thus, the radius of the loop is 91.84 m/s².
Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
