All categories

1 year ago

Which wavelength of light is the best choice when attempting to quantitatively relate solution absorbance and concentration?.

Physics

1 answer:

ANEK [815]1 year ago

6 0

The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)

you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.

<h3>What is wavelength ?</h3>

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).

The distance between two waves' crests serves as an illustration of wavelength. When you and another person have the same overall mindset and can easily communicate, that is an example of being on the same wavelength.

Learn more about Wavelength here:

brainly.com/question/10750459

#SPJ4

You might be interested in

(SCIENCE) 22 POINTS AND CROWN ASAP

gayaneshka [121]

Answer:

A or D

Explanation:

Net force includes addition or subtraction.

5 0

2 years ago

A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?

Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

$C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2$

∴ Area of each plate is 7.9060 m²

Voltage

$V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts$

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0

3 years ago

If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?

cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

3 0

2 years ago

What kind of force can be created with an electrical current?

RideAnS [48]

Answer:

magnatic force can be created

4 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

2 years ago