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3 years ago

8

10 kg box accelerates at 2 meters per second as it slides down a ramp at an angle of 25 degrees. What is the coefficient of fric

tion

1 answer:

maria [59]3 years ago

3 0

Answer:

0.241

Explanation:

resolving weight into two components and calculating force of friction in terms of coefficient of friction and then applying Newton 's law we get the value .This all has been explained in attachment

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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

So

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

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

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

A 0.16 kg billiard ball moving at 1.20 m/s collides elastically with one of the same mass at rest. It continues at a speed of 0.

Rainbow [258]

Use energy conservation to calculate the speed!
Total kinetic energy before collision = total kinetic energy after the collision.

5 0

3 years ago

An object example to decrease pressure.

Olin [163]

Answer:

To reduce pressure - decrease the force or increase the area the force acts on. If you were standing on a frozen lake and the ice started to crack you could lie down to increase the area in contact with the ice. The same force (your weight) would apply, spread over a larger area, so the pressure would reduce.

3 0

4 years ago

Read 2 more answers

A uniform cylinder of mass 1.5 kg and radius 0.3 m rolls down a ramp inclined at an angle 0.12 radians to the horizontal. What i

marishachu [46]

Answer:

<h3> different crystals from 6 to the point of the day I am not sure if you are aware of this but I am not sure if you are aware of this but I am not sure if you have any idea of how 3 is going to be able or not 2 3 or even if you have a good day or two to get to know you a 3 of a good place To Go for your own business or business 5 52 you are a great 49,494 548 4.</h3>

6 0

3 years ago

The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of

Fed [463]

Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is:

$\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)$

$\Delta l = 0.02\,m$

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

4 0

3 years ago