Answers:
1) 
2) 
Explanation:
1) Acceleration
is defined as the variation of Velocity
in time
:
(1)
A body also has acceleration when it changes its direction.
In this case we have a bus with a velocity of 60m/s to the east, that accelerates in a time 10s. So, we have to find the bus's acceleration:
(2)
(3) This is the bus's accelerration
2) Now we have a car that accelerates
to the west in order to reach a speed of
in the same direction, and we have to find the time
it takes to the car to reach that velocity.
Therefore we have to find
from (1):
(4)
(5)
Finally:
(6)
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>
<em></em>
<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>
<u><em></em></u>
<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>
Nicholas Copernicus correctly assumed that the planets revolved around the sun but he incorrectly assumed that the planets followed a perfect circle orbit around the sun. It was later on discovered by Johannes Kepler that the planets moved around the sun following an elliptical orbit.
Answer: A,B, and E
Explanation: Just checked I got them right:)
Answer:
5.3×10⁴ m/s
Explanation:
From the question,
Momentum = mass× velocity
M = mV................ Equation 1
Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane
make V the subject of the equation
V = M/m.................. Equation 2
Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg
Substitute these values into equation 2
V = 1.6×10⁹/3.0×10⁴
V = 5.3×10⁴ m/s