<span>Work is classified as anything the exerts energy from a system. Therefore, during the entire activity as Tina both holds the rope and actively pulls on the rope with her puppy she is completing work. Tina also does work when she walks toward and way from her puppy.</span>
Answer:
Explanation:
stiffness k = 160
m = 10
angular frequency ω = 
= 
= 4
ω = 4
Let x = 4 - A sinωt
when t = 0
x = 4 in
when t = 2 s , x = - 4
- 4 = 4 - A sinωt
8 = A sin 4 x 2
8 = A sin8
A = 8 / sin 8
= 8 / .989
= 8.09 in .
x = 4 - A sinωt
dx / dt = - Aω cosωt
v = - Aω cosωt
for t = 0
v = - Aω
= - 8.09 x 4
= - 32.36 in / s
initial velocity v = - 32.36 in /s
displacement x for t = 4s
x = 4 - 8.09 sin 4 x 4
= 4 - 8.09 sin 16
= 4 - 8.09 x - .2879
= 4 + 2.33
= 6.33 in.
c ) Amplitude of vibration A = 8.09 in .as calculated above .
Explanation:What is centripetal acceleration?
Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.
Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.
The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration
a
c
a
c
a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.
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
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
For ideal gas.
Answer:
A
Explanation:
a statement that can be tested through the scientific method
