Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²

x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K

K= 58.66 lb/ft
The damper coefficient for critically damped system


Cc= 12.7 lb.sec/ft
Answer:

Explanation:
To solve this problem we use the expression for the temperature film

Then, we have to compute the Reynolds number

Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number

but we also now that

but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is

In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
Answer:
450,000m = 450km = 4.5E5
32,600,000W = 32.6MW = 3.26E7
59,700,000,000cal = 59.7Gcal = 5.97E10
0.000000083s = 83ns = 8.3E-8
35,000Ω = 35kΩ = 3.5E4
Explanation:
Giga = 1,000,000,000
Mega = 1,000,000
kilo = 1,000
unit = 1
deci = .1
centi = .01
milli = .001
micro = .000001
nano = .0000000001
pico = .000000000001
You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.