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irina1246 [14]
3 years ago
7

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s

mall patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1 cm across, and you estimate that the distance from the window shade to the wall is about 4 m.
Estimate:
a. The average wavelength of the sunlight (in nm)
b. The diameter of the pinhole (in mm).
Physics
1 answer:
baherus [9]3 years ago
7 0

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}

Put the value into the formula

\lambda_{avg}=\dfrac{400+700}{2}

\lambda_{avg}=550\ nm

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

w=\dfrac{2.44\lambda L}{D}

D=\dfrac{2.44\lambda L}{w}

Put the value into the formula

D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}

D=0.537\ mm

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

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masya89 [10]

The statement 'diverse crops can become vulnerable to pests, disease, and climate change' is considered false regarding this last point (climate change).

<h3>What is adaptation?</h3>

Adaptation is an evolutionary process where a species fits certain environmental conditions.

Plant species are intrinsically vulnerable to climate change because they are adapted to certain climate (temperature) conditions.

In conclusion, the statement 'diverse crops can become vulnerable to pests, disease, and climate change' is considered false regarding this last point (climate change).

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2 years ago
Can a 12 inch rubber band fit a 22 pound rubber band ball
Vlada [557]

Answer:

if it is a thiner one, yes it will fit because they can expand a pretty far distance

Explanation:

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Multiple Choice
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The Atlantic Ocean provides the Maritime (Sea) Polar (Cold) air mass
You sure this isn't your Geography question?

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3 years ago
The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency diffe
mel-nik [20]

Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?

B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?

Answer:

The frequency change percentage is 9.94%

Explanation:

The frequency shift can be calculated as follows.

F= \frac{V+V_{0}}{V+V_{s}}

   = 9.05GHz\times\frac{343m/s+0}{343m/s+(-31.3m/s)}

   =9.95 GHz

So the frequency change seen by the detector is 9.95 - 9.05

% difference= \frac{0.9}{9.05} \times100

                    = 9.94%

3 0
3 years ago
At some point in an underground horizontal pipe, the speed of water is 2.5m/s and the gauge pressure is 2×104 Pa. What is the ga
lys-0071 [83]

The gauge pressure of the pipe is  2.234 ×10⁴  Pa, if the pipe is changed for another twice the area.

To find the gauge pressure the values given are,

speed of water = 2.5 m/s

Gauge pressure = 2×10⁴  Pa

<h3>What is pressure?</h3>

Definition:

      Pressure can be defined as the total applied force per unit of area. The pressure depends both on externally applied force as well the area on which it is applied.

To find the gauge pressure first we have use continuity equation,

A ₁V₁ = A₂V₂

By the given condition: As the area is increased to twice the original area

A ₁V₁ = 2A₁V₂

A₁ gets cancelled on both the sides,

V₂ =V₁/2

    =2.5 /2

V₂=1.25 m/s

Then, substituting the values in Bernoulli's equation,

Equation is given as,

P₁ + 1/2 ρV₁² = P₂ + 1/2 ρV₂²

Taking 1/2 ρ as common,

P₂ =  2×10⁴ + 1/2 ×1000×(2.5² -1.25²)

    = 2×10⁴ + 2343.75

    =22343.75 Pa

    =2.234 ×10⁴ Pa

Thus, if the pipe is changed for another twice the area then the gauge pressure is 2.234 ×10⁴ Pa.

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6 0
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