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Lena [83]
3 years ago
14

Design at least two independent experiments to determine the wavelength of a microwave source. You have a microwave source, micr

owave detector, a barrier with different regions, and a goniometer to measure angles. (To avoid over-ranging the detector, please use the largest meter multiplier setting that allows you to see what you’re doing. Nevertheless, if you are seeing nothing at any angle, do go to a smaller multiplier.) One of your experiments should involve setting up a standing wave.
One of your experiments should involve Include the following in your report:

a. Design experiments to solve the problem and discuss how you will use the available equipment to make measurements.
b. Describe the mathematical procedures you will use.
c. List the assumptions are you making. Explain how each could affect the outcome.
d. What are the sources of experimental uncertainty? How could you minimize the uncertainties?
Physics
1 answer:
Dominik [7]3 years ago
7 0

Answer:

The Double Slits Experiment and Michelson Interferometer

The questions will be answered for the double slits experiment.

(b) Mathematically, the double slits experiment equation can be given as: d sin 0 = m(Wavelength). d is the separation distance, 0 is the diffraction angle, m = 1,2,3,...

(c) assumptions

The width of the slits is lesser than the microwave's wavelength. This is to set up a standing wave between the microwave source and detector.

m is an positive integer. To obtain a constructive interference of the E-M wave(microwave)

(d) The uncertainties are:

(i)The zero error in the reading of the multiplier will disrupt the value of the wavelength by small percentage. This can be adjusted to obtain more accurate result.

(ii) The angle as obtained by the gionometer can not be measured to highest level of accuracy, as there are some approximations. High sensitive equipment should be used to obtain accurate result

Explanation:

Double Slits Experiment.

The double slits is a pair of 2cm wide slits with a 6cm separation, cut in a metal foil, with the double slits at the centre of turnable with the microwave source and microwave detector through the slits. The shunt on the microammeter is adjusted to

so that a large scale deflection is obtained. The interference pattern may then be explored by moving the receiver arm, whilst keeping the transmitter arm fixed. A graph of intensity (current) versus θ(Measured through the gionometer) should be plotted, and an estimate made of the microwave wavelength by measuring the angular separation of adjacent maxima in the interference pattern and the separation of the two slits.

Michelson Interferometer

Setting up the interferometer, the mirrors are metal sheets and the beamsplitter is a hardboard sheet. The metal sheets should be carefully positioned to be perpendicular to the microwave beam, and the beamsplitter positioned at the centre of the turntable and at 45° to the beam. Any slight change in the path length of either of the beams leaving the beamsplitter changes the interference pattern at the receiver. One of the metal sheets is mounted in a frame which slides along rails. If this sheet is slowly moved, maxima and minima of current will be recorded by the receiver and a plot of “intensity” against distance can be made. The distance between successive maxima (or minima) is one half of the microwave wavelength which can thus be measured.

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What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
Nikolay [14]

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

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
A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
3 years ago
If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add
mr Goodwill [35]

Each Al^+^3 ion contains three extra protons. Hence, the extra charge on each  Al^+^3 = 3 \times 1.6 \times 10^-^1^9 C

Total charge = 0.035 pC

Total charge (Q) = 0.035 \times 10^-^1^2 C

Let the number of Al^+^3 ions be n.

According to question:

n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2

n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}

n = 7.29167 \times 10^4

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0
3 years ago
A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
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