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Oksana_A [137]
3 years ago
8

A researcher measured the absorbance and % transmittance of a blue dye at light wavelengths between 425 and 700 nm. the measurem

ents were made every 25 nm in a spectrophotomter. how should she plot her data so she can determine the wavelengths of maximum absorbance and % transmittance?
Physics
2 answers:
Salsk061 [2.6K]3 years ago
7 0

Answer;

Plot the absorbance and the % transmittance on the y-axis, and the wavelength in nm on the x-axis.

Explanation;

Many compounds absorb ultraviolet (UV) or visible (Vis.) light. The amount of radiation absorbed may be measured in a number of ways:

  • Transmittance, T = P / P0
  • % Transmittance, %T = 100 T
  • Absorbance,  A = log10 P0 / P;
  • etc.

In this case; the wavelengths of maximum absorbance and % transmittance can be determined by Plotting the absorbance and the % transmittance on the y-axis, and the wavelength in nm on the x-axis.

MA_775_DIABLO [31]3 years ago
4 0
She should plot her <span>data where the y axis contains the absorbance values while the x-axis are the wavelengths. It should be that the dependent variable is always at the y-axis and the independent variable or the measurement that is manipulated is at the y-axis. Hope this helps.</span>
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Please help ASAP.
Elden [556K]

Answer:

1st question c part

2nd question c part

8 0
3 years ago
A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam
ahrayia [7]

Answer:

  1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

  (1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0
3 years ago
Read 2 more answers
HELP ASAP!!!
butalik [34]

Answer:

<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards </em>

Explanation:

<u>Vertical Throw </u>

When an object is thrown upwards, it describes a special type of motion ruled only by gravity.  

When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.

The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.  

With all these concepts in mind, we state that:

<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards  </em>

The other options are not correct   because:

A. The acceleration is never upwards  

B. The acceleration is never 0  

C. Both times are equal

8 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
6. The four main systems of the Earth are
nevsk [136]
Geosphere, hydrosphere, atmosphere, and biosphere.

Therefore, the correct answer is D.
8 0
3 years ago
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