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3 years ago

Describe ICP/OES in detail.

Engineering

2 answers:

alex41 [277]3 years ago

7 0

Answer:

ICP -OES stand for inductively coupled plasma optical emission spectroscopy

Explanation:

It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.

The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.

makkiz [27]3 years ago

6 0

Answer:

ICP stands for Insane Clown Posse. It is a rap group. The group's members are Violent J and Shaggy 2 Do*e

Explanation:

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A jet of water 75m in diameter,issues with a velocity of 30m/s and impinge on a stationary plate which distort its forward motio

pashok25 [27]

Forces exerted by the jet on the plate is=3976N

<h3>How to calculate forces exerted by the jet on the plate?</h3>

A force is an effect that can change the motion of an object. A force can generate an object with mass to change its velocity, i.e., to accelerate. Force can also be explained intuitively as a push or a pull. A force has both volume and direction, making it a vector amount.

A jet of water 75m in diameter

velocity = 30m/s

The forces exerted by the jet on the plate is

F=1000×44178×10^-3×30²

=3976N

the jet on the plate work done by Is zero .

To learn more about Force, refer

brainly.com/question/12970081

#SPJ9

5 0

1 year ago

One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p

kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0

3 years ago

Write a grammar for a language whose sentences start with an even and non-zero number of x’s, end with an odd number of z’s, and

sveta [45]

Answer:

GRAMMAR

S -> AB

A -> xxAyy | xxyy

B -> yyBzz | yz

EXPLANATION

A is for even number of x's followed by that number of y's

B is for odd number of y's followed by that number of z's

3 0

3 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and gauge pressure of 300 kPa. The gas is h

Sergio [31]

Answer:

the final temperature is 77.1 °C

Explanation:

Given the data in the question;

Initial temperature; T₁ = 27°C = ( 27 + 273)K = 300 K

Initial absolute pressure P₁ = 300 kPa = ( 300 + 101.325 )kPa = 401.325 kPa

Final absolute pressure P₂ = 367 kPa = ( 367 + 101.325 )kPa = 468.325 kPa

Now, to calculate the final temperature, we use the ideal gas equation;

P₁V/T₁ = P₂V/T₂

but it is mentioned that the rigid tank is closed,

so the volume is the same both before and after.

Change in volume = 0

hence;

P₁/T₁ = P₂/T₂

we substitute

401.325 kPa / 300 K = 468.325 kPa / T₂

T₂ × 401.325 kPa = 300 K × 468.325 kPa

T₂ = [ 300 K × 468.325 kPa ] / 401.325 kPa

T₂ = 140497.5 K / 401.325

T₂ = 350.08 K

T₂ = ( 350.08 - 273 ) °C

T₂ = 77.1 °C

Therefore, the final temperature is 77.1 °C

3 0

3 years ago

Question Completion Status:

Mashutka [201]

Answer: c. Centre of pressure

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

6 0

2 years ago