Answer:
the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour
Explanation:
the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is
V = πR² * h
thus
h = V / (πR²)
Considering that the volume of the slick remains constant, the rate of change of radius will be
dh/dt = V d[1/(πR²)]/dt
dh/dt = (V/π) (-2)/R³ *dR/dt
therefore
dR/dt = (-dh/dt)* (R³/2) * (π/V)
where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness
when the radius is R=8 m , dR/dt is
dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour
<u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u>
Explanation:
<h2>Given:</h2>
Acceptable volume = 40,000 cm³
Area of luggage = 1,960 cm²
Height of luggage = 23 cm
<h2>Question:</h2>
Is the passenger's luggeage ok to carry onto the airplane
<h2>Equation:</h2>
V = l x w x h
or we can use
V = A x h
Since A = l x w
where: V - volume
A - area
l - length
w - width
h - height
<h2>Solution:</h2>
V = A x h
V = ( 1,960 cm²)(23 cm)
V = 45,080 cm³
45,080 cm³ is greater than the acceptable volume 40,000 cm³
<h2>Final Answer:</h2><h3><u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u></h3><h3 />
Hope that helps had to do it myself http://160592857366.free.fr/joe/ebooks/Mechanical%20Engineering%20Books%20Collection/FLUID%20MECHANICS/Fundamentals%20of%20Fluid%20Mechanics%204th%20Edition%20-%20Munson%20-%20John%20Wiley%20and%20Sons/MUNSON%20-%20livro.pdf
