Answer:
Physical equilibrium is defined as the equilibrium which develops between different phases or physical properties. In these processes, there is no change in chemical composition. The following are some examples of Physical Equilibria: Solid-Liquid Equilibria. Please give me the brainliest answer?
:) Hoped this helped!!! Have a good day!!! <3
<span>This phenomenon is known as illusion of attention and sometimes can include inattentional blindness. Basically, as humans, we are not used to backgrounds and people suddenly changing so our brains are not thinking to look for it.</span>
Answer:
Tension.
Explanation:
The tighter a string is, the more tension is applied to it; the more loose a string is, less tension is applied to it.
Therefore, when you adjust the tightness of a string you are tackling with it's tension.
Hope it helped,
BiologiaMagister
<u>Answer:
</u>
Average speed, over the first 4.0 s of its motion, of a pebble released from rest off a bridge = 19.6 m/s
<u>Explanation:
</u>
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case the initial velocity of body = 0 m/s, acceleration = 9.8
, we need to calculate displacement when t = 4 seconds.
Substituting
![S=0*t+\frac{1}{2} *9.8*4^2\\ \\ S=78.4 meter](https://tex.z-dn.net/?f=S%3D0%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.8%2A4%5E2%5C%5C%20%5C%5C%20S%3D78.4%20meter)
So a distance of 78.4 meter has been traveled by pebble in 4 seconds.
Average speed = 78.4/4 = 19.6 m/s
Answer:
Transmissivity is ![186.96\times 10^6 ft^2/sec](https://tex.z-dn.net/?f=186.96%5Ctimes%2010%5E6%20ft%5E2%2Fsec)
Explanation:
So the well 1 has
r1=26ft
h1=29.34ft
The well 2 has
r2=73 ft
h2=32.56 ft
Converting the rate of flow from gal/min to ft3/day
![Q=220\frac{gal}{min}\times\frac{1 ft^3}{7.48 gal}\times\frac{1440 min}{1 day}\\Q=42400\frac{ft^3}{day}](https://tex.z-dn.net/?f=Q%3D220%5Cfrac%7Bgal%7D%7Bmin%7D%5Ctimes%5Cfrac%7B1%20ft%5E3%7D%7B7.48%20gal%7D%5Ctimes%5Cfrac%7B1440%20min%7D%7B1%20day%7D%5C%5CQ%3D42400%5Cfrac%7Bft%5E3%7D%7Bday%7D)
Now the transmissivity is is given as
![T=\frac{\dot{Q}}{2\pi(h_2-h_1)}ln(\frac{r_2}{r_1})](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B%5Cdot%7BQ%7D%7D%7B2%5Cpi%28h_2-h_1%29%7Dln%28%5Cfrac%7Br_2%7D%7Br_1%7D%29)
Now by substituting values
![T=\frac{\dot{Q}}{2\pi(h_2-h_1)}ln(\frac{r_2}{r_1})\\T=\frac{42400}{2\pi(32.56-29.34)}ln(\frac{73}{26})\\T=\frac{42400}{6.44\pi }\left(\ln \left(73\right)-\ln \left(26\right)\right)\\T=2164 ft^2/day](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B%5Cdot%7BQ%7D%7D%7B2%5Cpi%28h_2-h_1%29%7Dln%28%5Cfrac%7Br_2%7D%7Br_1%7D%29%5C%5CT%3D%5Cfrac%7B42400%7D%7B2%5Cpi%2832.56-29.34%29%7Dln%28%5Cfrac%7B73%7D%7B26%7D%29%5C%5CT%3D%5Cfrac%7B42400%7D%7B6.44%5Cpi%20%7D%5Cleft%28%5Cln%20%5Cleft%2873%5Cright%29-%5Cln%20%5Cleft%2826%5Cright%29%5Cright%29%5C%5CT%3D2164%20ft%5E2%2Fday)
Now converting it to ft^2/sec
![T=2164 ft^2/day\\T=186.96\times 10^6 ft^2/sec](https://tex.z-dn.net/?f=T%3D2164%20ft%5E2%2Fday%5C%5CT%3D186.96%5Ctimes%2010%5E6%20ft%5E2%2Fsec)
Transmissivity is ![186.96\times 10^6 ft^2/sec](https://tex.z-dn.net/?f=186.96%5Ctimes%2010%5E6%20ft%5E2%2Fsec)