I found a similar problem from another website which shows the diagram. This is because we need to compute first the net force to determine the work. Afterwhich, we can use the formula: Work = Force*Distance.
Net Force = 12.2 N - wcosθ
Since the weight is 30 N and the angle is 24°,
Net Force = 12.2 - 30cos24° = -15.2 N
So,
W = (-15.2 N)(3.60 m)
<em>W = -54.7 J</em>
Dispersion-Distributing something over an area.Diffusion-the scattering of light by reflection from a rough surface.
Refraction-<span>Turning or bending of any wave.</span>
Reflection- Image.
Answer:
Explanation:
Magnetic field due to circular wire at the center = μ₀ I / 2 r
I is current and r is radius . μ₀ = 4π x 10⁻⁷.
field B₁ due to inner loop
B₁ = 4π x 10⁻⁷ x 12 / 2 x .20
= 376.8 x 10⁻⁷
Field due to outer loop
B₂ = 4π x 10⁻⁷ x I / 2 x .30
For equilibrium
B₁ = B₂
376.8 x 10⁻⁷ = 4π x 10⁻⁷ x I / 2 x .30
I = 18 A.
The direction should be opposite to that in the inner wire . It should be anti-clockwise.
Answer:
![\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp](https://tex.z-dn.net/?f=%5Cmu%20_j%3D%5Cdfrac%7B1%7D%7BC_p%7D%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
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
Pressure and temperature at inlet and
Pressure and temperature at exit
We know that Joule -Thompson coefficient given as

Now from T-ds equation
dh=Tds=vdp
So
![Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp](https://tex.z-dn.net/?f=Tds%3DC_pdt-%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p%5Cright%5Ddp)
⇒![dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp](https://tex.z-dn.net/?f=dh%3DC_pdt-%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
So Joule -Thompson coefficient
![\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp](https://tex.z-dn.net/?f=%5Cmu%20_j%3D%5Cdfrac%7B1%7D%7BC_p%7D%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
This is Joule -Thompson coefficient for all gas (real or ideal gas)
We know that for Ideal gas Pv=mRT

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](https://tex.z-dn.net/?f=%5Cmu%20_j%3D%5Cdfrac%7B1%7D%7BC_p%7D%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
For ideal gas.
Answer:
C. An inital volocity that is faster than the final volocity
Explanation:
.