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2 years ago

Hey can anyone please help me with this it’s due in few hours and I’m stuck

Physics

1 answer:

Margaret [11]2 years ago

7 0

Answer:

1) P = rho g h = d g h where d is replacing the greek letter rho (density)

a) pressure at A must be 1/2 that of B because of 1/2 the height

b) pressure at B equals that at D because the height is the same

(Pascal's principle - pressure in a fluid is same in all directions)

c) pressure A is the same as that C

d) the pressure at B would be .8 of the water at B if paraffin replaced the water (P = d g h = .8 * dw g h) due to change in density

2) Volume = 4 m X 3 m X 2 m) = 24 m^3 volume of tank

b) mass = volume * density = 24 m^3 * 800 kg / m^3 = 19,200 kg

c) weight = mass * gravity = 19,200 kg * 9.8 m/s^2 = 188,160 Newton's

d) P = d g h = 800 kg/m^3 * 9.8 m/s^2 * 2 m 15,680 N/m^2

Pressure is given in force/area or N/m^2

3) Pw / Pp = dw / dp = 1000 / 800 = 1.25

Pw = 1.25 Pp = 15,680 m/s^2 * 1.25 = 19,600 Pascal's

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

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