The answer is the last option "Respiration"
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
= 
So,
Rearranging the above equation to get Electric field, we will get:
E = 
Multiply and divide by
E =
x 
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x
) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
= 
Rearranging the above equation for E:
E = 
E =
+ 
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E =
+ 
Replacing the above equations to get E:
E = (σ1 x
) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x
= - σ2 x 
Answer:
The radius is 
Explanation:
From the question we are told that
The distance beneath the liquid is 
The refractive index of the liquid is 
Now the critical value is mathematically represented as
![\theta = sin ^{-1} [\frac{1}{n_i} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7Bn_i%7D%20%5D)
substituting values
![\theta = sin ^{-1} [\frac{1}{131} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B131%7D%20%5D)

Using SOHCAHTOA rule we have that

=> 
substituting values


Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
Answer:
distance - meters
speed - meters/seconds
time - seconds
velocity - meters/seconds
acceleration - meters/seconds²