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Answer:
a-1 Graph is attached. The relation is linear.
a-2 The corresponding height for 68 kPa Pressure is 7.54 m
a-3 The corresponding weight for 68 kPa Pressure is 1394726kg
b The original height of the column is 5.98 m
Explanation:
Part a
a-1
The graph is attached with the solution. The relation is linear as indicated by the line.
a-2
By the equation

Here
- P is the pressure which is given as 68 kPa.
- ρ is the density of the oil whose SG is 0.92. It is calculated as

- g is the gravitational constant whose value is 9.8 m/s^2
- h is the height which is to be calculated

So the height of column is 7.54m
a-3
By the relation of volume and density

Here
- ρ is the density of the oil which is 920 kg/m^3
- V is the volume of cylinder with diameter 16m calculated as follows

Mass is given as

So the mass of oil leading to 68kPa is 1394726kg
Part b
Pressure variation is given as

Now corrected pressure is as

Finding the value of height for this corrected pressure as

The original height of column is 5.98m
Explanation:
The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
You haven't said how much power the stereo uses. It matters !
Whatever that number is, the maximum hours per month is
(3460) divided by (the # of watts the stereo uses when it's playing) .