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dalvyx [7]
3 years ago
14

A ball is dropped from rest.

Physics
1 answer:
nignag [31]3 years ago
6 0

Answer:

How fast is it going? 29.4 Meters per second

How far has it fallen? 44.1 meters

Explanation: Gravitional Acceleration: 9.8 meters per secnd squared!

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How do you find motion
saveliy_v [14]

Answer:

idk you got me there

Explanation:

8 0
3 years ago
Burning a canful of petrol or dropping it. which is likely to release most energy?​
Neporo4naja [7]

Answer:

Burning of canful of petrol will release more energy

Explanation:

If you like my answer than please mark me brainliest

5 0
3 years ago
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morn
never [62]

Answer:

b) 7.00

Explanation:

N( t ) = -20( t - 5 )²

dN/ dt = -20 x 2 ( t - 5 )

For maximum N ( depth )

dN/dt = 0

- 40 ( t - 5 ) = 0

t = 5

So at 2 + 5 = 7 .00 am depth of water reaches its maximum.

6 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
A nucleus with a mass number of 64 has a mean radius of about: _________.
dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is A= 64

Generally the mean radius is mathematically evaluated as

R  =  R_o A^{\frac{1}{3} }

Here R_o is a constant with a value  R_o =1.2*10^{-15}

So  

     R  =  1.2*10^{-15} *  64^{\frac{1}{3} }

      R  =   4.8 *10^{-15}

        R  =   4.8\ fm

5 0
3 years ago
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