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kompoz [17]
3 years ago
15

a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike

s the ground is most nearly​
Physics
1 answer:
EleoNora [17]3 years ago
4 0

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

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Roller coasters work by utilizing its potential energy. Potential energy is "lost" as the cars lose height, subsequently gaining kinetic energy as revealed by increased speeds (and loop the loop). As the cars of the roller coaster course through the changing heights, it constantly swaps between potential and kinetic energy. Theoretically, this process could be endless. However, energy is continually "lost" because of dissipative forces such as friction and air resistance.

5 0
3 years ago
Do you think that rubbing two pieces of woods together would have more or less friction than wood rubbing against smooth metal ?
salantis [7]

Answer: More friction

Explanation: The wood would probably splinter, causing even more friction, while the metal would probably smooth the wood out.

7 0
3 years ago
A human hair has a thickness of about 70 micrometers. How many meters is this?
Aleksandr [31]
This is 7e-5 meters. Hope this helps! Please mark brainliest. :)
6 0
2 years ago
Read 2 more answers
Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont
SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

T = \frac{2 u Sin\theta }{g}

T = \frac{2 \times 20 \times Sin35 }{9.8}

T = 2.34 second

(b) The formula for the maximum height is given by

H = \frac{u^{2} \times Sin^{2}\theta }{2g}

H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}

H  = 6.71 m

(c) The formula for the range is given by

R = \frac{u^{2} \times Sin 2\theta }{g}

R = \frac{20^{2} \times Sin 2 \times 35}{9.8}

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0
3 years ago
An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-
maw [93]

Answer:

  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. 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
  4. 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:

  1. 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
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  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>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

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>

4 0
3 years ago
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