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soldier1979 [14.2K]

4 years ago

12

How has technology influenced theories over time?

1 answer:

Contact [7]4 years ago

7 0

Better technology is helping us because we can see more stuff like the microscope we able to make assumptions based on what we saw.

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A note of frequency 2000Hz has a velocity of 400 ms! What is the wavelength of the note?

Rudik [331]

Answer:

Explanation:

frequency=c/λ

c=400m/s

putting values

2000=400/λ

λ=0.2m

8 0

3 years ago

Read 2 more answers

What is the frequency of blue light that has a wavelength of 446 nm?

Maslowich

When I went through with the math, the answer I came upon was:
<span>6.67 X 10^14 </span>

<span>Here is how I did it: First of all we need to know the equation. </span>

<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>

<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>

<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>

<span>Answer: 6.67 X 10^14 s- hope this helps</span>

7 0

4 years ago

Read 2 more answers

What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes

sp2606 [1]

** Missing info: Lines per mm = 500 **

Ans: The wavelength is = λ = 1414.21 nm

Explanation:
The formula for diffraction grading is:

dsinθ = mλ --- (1)

Where
d = 1/lines-per-meter = (1/500)*10^-3 = 2 * 10^-6
m = order = 1
λ = wavelength
θ = 45°

Plug in the values in (1):
(1) => 2*10^-6*sin(45°) = (1)λ
=> λ = 1414.21 nm

7 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The sec

SpyIntel [72]

Answer:

2.32
0.43

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

$\frac{Vn}{Vo} = \frac{Nn}{No}$ = $\frac{730}{315}$ therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

$\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }$ equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

$\frac{In}{Io} = \frac{Vo}{Vn}$ = 315 / 730 = 0.43

7 0

3 years ago