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Pepsi [2]
3 years ago
5

What is the critical angle of light traveling from vegetable oil into water

Physics
1 answer:
Bad White [126]3 years ago
5 0

56.1∘

Question: A glass is half-full of water, with a layer of vegetable oil (n = 1.47) floating on top. A ray of light traveling downward through the oil is incident on the water at an angle of 56.1∘ .

A glass is half-full of water, with a layer of vegetable oil (n ...https://study.com › academy › answer › a-glass-is-half-ful...

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What is the average kinetic energy of an object's particles?

the answer is temperature

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Whose law states that an induced current will create a magnetic field?
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An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
kipiarov [429]

Answer:

2.24 \times 10^{-14} Newtons

Explanation:

Magnitude of charge on the electron = q = 1.6 \times 10^{-19} Coulombs

The negative sign in the question statement indicates that the charge is negative.

Magnitude of Electric Field experienced by the electron = E = 1.4 \times 10^{5} Newtons/Coulomb

Magnitude of Force on the electron = F = ?

The relation between the charge, electric field and the force on the charge because of electric field is given by:

E=\frac{F}{q}

From here we can write:

F = qE

Using the values, we get:

F = 1.6 \times 10^{-19} \times 1.4 \times 10^{5}\\\\ F = 2.24 \times 10^{-14}

Thus, the magnitude of the electric force experience by the electron would be 2.24 \times 10^{-14} Newtons

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
Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the
Anit [1.1K]

Answer:

C. Quadruple

Explanation:

¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

3 0
2 years ago
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