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Alex17521 [72]
3 years ago
7

A researcher measured the absorbance and percent transmittance of a blue dye at wavelengths between 425 and 700 nmnm. The measur

ements were made every 25 nmnm in a spectrophotometer. How should she plot her data to determine the wavelengths of maximum absorbance and percent transmittance?
Physics
1 answer:
Alina [70]3 years ago
7 0

Answer:

values ​​the transmittance or absorbance is plotted on the y axis, dependent variable and the wavelength on the x axis, independent variable.

Explanation:

The absorbance or transmittance measurements (T) determine the amount of light that passes through a material (I) and divide it by the amount of light that passes without the material (Io), to see the values ​​the transmittance or absorbance is plotted on the y axis, dependent variable and the wavelength on the x axis, independent variable.

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The role of learning in motivation is most obvious from the influence of<br><br>​
taurus [48]

A thing that motivates or encourages one to do something is an incentive. Answer: Incentives.

6 0
3 years ago
Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small
Hatshy [7]

Answer:

Electric Field a the centreE=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle \theta\\ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let \lambda be the charge per unit length such that\lambda=\dfrac{Q}{\pi R}

k=\dfrac{1}{4\pi \epsilon_0}

Total Electric Field at the centre

=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta\\

integrating 0 to \dfrac{\pi}{2}

E=\dfrac{2k\lambda}{R}(-\vec j)

E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

So the Electric field at the centre is calculated.

7 0
3 years ago
A positive charge of 6.0 x 10-4 C is in an electric field that exerts a force of 4.5 x 10 -4 on it. What is the strength of the
Yuki888 [10]

Electric field is defined as force per unit charge.

So it is given by

F = q E

now we can find electric field by

E = \frac{F}{q}

E = \frac{4.5*10^{-4}}{6 * 10^{-4}}

E = 0.75 N/C

So field strength is 0.75 N/C.

6 0
4 years ago
Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods c
EastWind [94]

Answer:

I think it will be half of the initial charge

Explanation:

Because we know, the resulting charge will be q1+q2/2, since one is neutral so the charge will be half q/2

3 0
3 years ago
A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd
Delicious77 [7]

Answer:\theta =\tan ^{-1}(\frac{v^2}{gR})

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

R\cos \theta will balance weight and R\sin \theta will provide the necessary centripetal Force

thus R\sin \theta =\frac{mv^2}{R}  ------------1

R\cos \theta =m g  ----------------2

Divide 1 & 2 we get

\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}

\tan \theta =\frac{v^2}{gR}

\theta =\tan ^{-1}(\frac{v^2}{gR})                                      

8 0
3 years ago
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