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4 years ago

What is the UL measurement?

1 answer:

4 0

The UL measurement is a symbol used by the International System of Units to represent a microliter. A microliter is a metric unit of measurement for liquid volume and is equal to 1/1,000,000 (one-millionth) of a liter.

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What is the formula for Charles Law?

Vlada [557]

Answer:

Explanation:

V_{1} = first volume

V_{2} = second volume

T_{1} = first temperature

T_{2} = second temperature

4 0

3 years ago

Experiments show that the pressure drop for flow through an orifice plate of diameter d mounted in a length of pipe of diameter

Klio2033 [76]

The question is not clear and the complete question says;

Experiments show that the pressure drop for flow through an orifice plate of diameter d mounted in a length of pipe of diameter D may be expressed as Δp = p1 − p2 =f (ρ, μ, V, d, D). You are asked to organize some experimental data. Obtain the resulting dimensionless parameters.

Answer:

The set of dimensionless parameters is; (Δp•d)/Vµ = Φ((D/d), (ρ•d•V/µ))

Explanation:

First of all, let's write the functional equation that lists all the variables in the question ;

Δp = f(d, D, V, ρ, µ)

Now, since the question said we should express as a suitable set of dimensionless parameters, thus, let's write all these terms using the FLT (Force Length Time) system of units expression.

Thus;

Δp = Force/Area = F/L²

d = Diameter = L

D = Diameter = L

V = Velocity = L/T

ρ = Density = kg/m³ = (F/LT^(-2)) ÷ L³ = FT²/L⁴

µ = viscosity = N.s/m² = FT/L²

From the above, we see that all three basic dimensions F,L & T are required to define the six variables.

Thus, from the Buckingham pi theorem, k - r = 6 - 3 = 3.

Thus, 3 pi terms will be needed.

Let's now try to select 3 repeating variables.

From the derivations we got, it's clear that d, D, V and µ are dimensionally independent since each one contains a basic dimension not included in the others. But in this case, let's pick 3 and I'll pick d, V and µ as the 3 repeating variables.

Thus:

π1 = Δp•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π1 = F/L²•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π1 to be dimensionless,

π1 = F^(0)•L^(0)•T^(0)

Thus;

F/L²•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F,

1 + c = 0 and c= - 1

For L; -2 + a + b - 2c = 0

For T; -b + c = 0 and since c=-1

-b - 1 = 0 ; b= -1

For L, -2 + a - 1 - 2(-1) = 0 ; a=1

So,a = 1 ; b = -1; c = -1

Thus, plugging in these values, we have;

π1 = Δp•d^(1)•V^(-1)•µ^(-1)

π1 = (Δp•d)/Vµ

Let's now repeat the procedure for the second non-repeating variable D2.

π2 = D•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π1 = L•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π2 to be dimensionless,

π2 = F^(0)•L^(0)•T^(0)

Thus;

L•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F;

-2c = 0 and so, c=0

For L;

1 + a + b - 2c = 0

For T;

-b + c = 0

Since c =0 then b =0

For, L;

1 + a + 0 - 0 = 0 so, a = -1

Thus, plugging in these values, we have;

π2 = D•d^(-1)•V^(0)•µ^(0)

π2 = D/d

Let's now repeat the procedure for the third non-repeating variable ρ.

π3 = ρ•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π3 = F/T²L⁴•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π4 to be dimensionless,

π3 = F^(0)•L^(0)•T^(0)

Thus;

FT²/L⁴•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F;

1 + c = 0 and so, c=-1

For L;

-4 + a + b - 2c = 0

For T;

2 - b + c = 0

Since c =-1 then b = 1

For, L;

-4 + a + 1 +2 = 0 ;so, a = 1

Thus, plugging in these values, we have;

π3 = ρ•d^(1)•V^(1)•µ^(-1)

π3 = ρ•d•V/µ

Now, let's express the results of the dimensionless analysis in the form of;

π1 = Φ(π2, π3)

Thus;

(Δp•d)/Vµ = Φ((D/d), (ρ•d•V/µ))

3 0

3 years ago

What do we call point E

vazorg [7]

Focal point of reflection

5 0

3 years ago

Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh

ahrayia [7]

Answer:

i) FALSE, ii) TRUE, iii) FALSE, iv) FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

nₓ <n_y

i) The expression of the refractive index

nₓ < n_y

$\frac{c}{v_x} = \frac{c}{n_y}$

v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

the statement of FALSE

8 0

3 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel $N_1=2\ rev/s$

angular speed $\omega_1=2\pi N_1=4\pi\ rad/s$

mass of wheel $m_1=4.5\ kg$

diameter of wheel $d_1=0.30\ m=30\ cm$

radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

Moment of inertia of Wheel

$I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2$

Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$

Common frequency of wheel and chunk of clay is

$\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s$

5 0

3 years ago