Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
Answer:
conditional instability (Γd > Γe > Γw)
Explanation:
Given;
dry adiabatic rate, Γd = 10ºC/1000 meters
wet adiabatic rate, Γw= 6.5ºC/1000 meters
environmental lapse rate, Γe = 7.8ºC/1000 meters
Stability of the atmosphere can be described as Absolute stability, Absolute instability or conditional instability.
Conditions for Absolute stability:
Γd > Γw > Γe
Conditions for Absolute instability:
Γe > Γd > Γw
Conditions for conditional instability:
Γd > Γe > Γw
Thus, conditional instability satisfies the given values of the atmospheric condition: Γd (10) > Γe (7.8) > Γw (6.5)
