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3 years ago

The statement “the scientific process is open-ended “means

Physics

2 answers:

nordsb [41]3 years ago

6 0

It is a question that is answered with a statement, not just a yes or no

Alex73 [517]3 years ago

6 0

Answer: The problem is the scientific question to be solved. It is best expressed as an "open- ended" question, which is a question that is answered with a statement, not just a yes or a no.

Explanation:

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Addition of _____________ to pure water causes the least increase in conductivity.

Mademuasel [1]

Answer:

salt

Explanation:

5 0

3 years ago

Discuss Joule-Thompson effect with relevant examples and formulae.

Delicious77 [7]

Answer:

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

Explanation:

Joule -Thompson effect

Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process

Let

$P_1,T_1$ Pressure and temperature at inlet and

$P_2,T_2$ Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

$\mu _j=\left(\frac{\partial T}{\partial p}\right)_h$

Now from T-ds equation

dh=Tds=vdp

$Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp$

⇒ $dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

So Joule -Thompson coefficient

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

$\dfrac{\partial v}{\partial T}=\dfrac{v}{T}$

So by putting the values in

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

$\mu _j=0$ For ideal gas.

6 0

3 years ago

Please help!<br> If an object has a mass of 500 g and floats, what volume of water must it displace?

mafiozo [28]

It must displace at least 500 milliliters (0.5 liter) of water, in order to float in water.

4 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$