Answer:
0.61 second.
Explanation:
So, from the question we are given the following useful parameters or data or information that is going to assist us in solving this question and they are;
(1). Inclination of plane, θ = 50°.
(2). ''The horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane" that is height = 0.92 m(92cm).
(3). The "acceleration of the cylinder to be 2/3 × g Sin (theta) and g = Acceleration due to gravity = 9.8m/s^2
Soz let us delve right into the solution to the question above. We will be making use of the mathematical representation or Equation below;
s = (1/2) a × t².
Where s = height, a = acceleration due to gravity and t = time.
0.92 = (9.8 × sin 50°/ 3) × t².
t² = 0.37
t = √ 0.37.
t =0.61 seconds.
This statement about the potential due to the sphere is correct: THE POTENTIAL IS HIGHEST AT THE CENTER OF THE SPHERE.
The potential is highest at the center of the sphere because, charge usually flow from the region of higher potential to lower potential.
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:
Explanation:
i. CW moment = 10 N (10 cm) + 30 N (30 cm) - 60 N (40 cm) = - 1400 N-cm
ii. ACW momenet = 60 N (40 cm) - 10 N (10 cm) + 30 N (30 cm) = 1400 N-cm
iii. No. The lever is not balanced in the situation. Because the moment is ± 1400 N-cm. if balance, the moment must be Zero.
iv. the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm
take moment from Δ (support)
60(40) = 10(x) + 30(30)
2400 = 10x + 900
10x = 2400 - 900
10x = 1500
x = 1500/10
x = 150 cm
therefore, the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm
Answer:
Chemical etching is a process of printed circuit board (PCB) manufacturing that provides many advantages over mechanical methods.
Explanation:
There's one! UvU hope this helped in whatever you're in for! <3
Good Luck!