Number of barrels are 3.0. Each barrel contains 42 gallons of oil. Thus, total volume of oil will be 42×3=126 gallons.
Converting gallons into m^{3}
1 gallon=0.00378 m^{3}
Thus, 126 gallons=0.4769 m^{3}
Thickness of oil film is 2.5\times 10^{2} nm, converting it into meters as follows:
1 nm=10^{-9} m
Thus,
2.5\times 10^{2} nm=1.5\times 10^{-7}m
Now, volume V of oil is related to area A and thickness T as follows:
V=A×T
rearranging,
A=\frac{V}{T}=\frac{0.4769 m^{3}}{2\times 10^{-7}m}=2.38\times 10^{6}m^{2}
Thus, square meters of oil will be 2.38\times 10^{6}m^{2}
Answer:
t = 2 v₀ / g
Explanation:
For this projectile launch exercise we use the displacement equations
x = vox t
y = y₀ + 
t - ½ g t²
As it is launched horizontally the vertical velocity is zero and the point of origin of the coordinate system is here, so y₀ is zero.
x = v₀ t
y = ½ g t²
They ask us for the time for which
x = y
vo t = ½ g t²
t = 2 v₀ / g
Answer:
To convert a millisecond measurement to a second measurement, divide the time by the conversion ratio. The time in seconds is equal to the milliseconds divided by 1,000.
Explanation:
hope it helps
The qualifications boil down to: College education.
In most university or industrial research organizations, you might be able to work there as a member of the team who doesn't get much pay or much respect, with research going on all around you directed by other people, after you've gotten you Master's degree.
But you really don't have a shot at leading anything, or having much to say about what's being researched or how, until you have a PhD degree in the field where you'd like to do the research.
(Did I mention how proud I was to be present about 6 weeks ago, in a land far away, when my daughter was awarded a PhD degree in Molecular Biology ? I didn't want to let you get away without hearing about that.)
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°
