Teams A and B are in a tug-of-war challenge. Team A wins the challenge.
2 answers:
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<h2>Answer: 469 feet</h2>
Explanation:
This problem is a good example of Vertical motion, where the main equation for this situation is:
(1)
Where:
is the height of the stone at 6s (the value we want to find)
is the initial height of the stone
is the initial velocity of the stone
is the time at which we need to find the height
is the acceleration due to gravity
Having this clear, let's find from (1):
(2)
Finally:
This is the height of the stone at t=6s
Answer:
- Power requirement <u>P</u> for the banner is found to be 30.62 W
- Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
- Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
- Power requirement <u>P</u> for the smooth spherical balloon was found to be 40.08 W
Explanation:
First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:
- v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
- The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
- The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
- The equation to determine drag-force is:
In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>
Frontal area A of the banner is : 25 x 0.8 = 20 m^2
<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>
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<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28
Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>
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<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption
<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:
Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N
P = 0.962 * 41.66 = <u><em>40.08 W</em></u>