Answer:
4.2·10⁷
Explanation:
Exponents are added when multiplying:
(a^b)(a^c) = a^(b+c)
1.15·10⁵ × 3.65·10² ≈ 4.2·10⁷
_____
Your calculator is designed to work with scientific notation. Scientific or graphing calculators can also display their output in scientific notation. A spreadsheet can also perform this calculation for you and display the result in the required format.
Calculators and spreadsheets sometimes use "E" in place of "×10^". That is, 4.2E7 means 4.2×10^7.
Answer:
V=1601gal
Explanation:
Hello! This problem is solved as follows,
First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.
This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.
P=Poil+Patm
P=total pressure or absolute pressure=26psi=179213.28Pa
Patm= the atmospheric pressure =101325Pa
Poil=pressure due to the weight of olive oil=0.86αgh
α=density of water=1000kg/m^3
g=gravity=9.81m/s^2
h= height that olive oil reaches
solving
P=Poil+Patm
P=Patm+0.86αgh
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Now we can use the equation that defines the volume of a cylinder.
V=
D=3ft=0.9144m
h=9.23m
solving
finally we use conversion factors to find the volume in gallons
Answer:
equivalent stiffness is 136906.78 N/m
damping constant is 718.96 N.s/m
Explanation:
given data
mass = 120 kg
amplitude = 120 N
frequency = 320 r/min
displacement = 5 mm
to find out
equivalent stiffness and damping
solution
we will apply here frequency formula that is
frequency ω = ω(n) √(1-∈ ²) ......................1
here ω(n) is natural frequency i.e = √(k/m)
so from equation 1
320×2π/60 = √(k/120) × √(1-2∈²)
k × ( 1 - 2∈²) = 33.51² ×120
k × ( 1 - 2∈²) = 134752.99 .....................2
and here amplitude ( max ) of displacement is express as
displacement = force / k × ( )
put here value
0.005 = 120/k × ( )
k ×∈ × √(1-2∈²) = 1200 ......................3
so by equation 3 and 2
solve it and we get
∈ = 1.00396
and
∈ = 0.08869
here small value we will consider so
by equation 2 we get
k × ( 1 - 2(0.08869)²) = 134752.99
k = 136906.78 N/m
so equivalent stiffness is 136906.78 N/m
and
damping is express as
damping = 2∈ √(mk)
put here all value
damping = 2(0.08869) √(120×136906.78)
so damping constant is 718.96 N.s/m