Answer:
3.222 ohms
Explanation:
If the total wire had a resistance of 29 ohms, when cut in three, each piece will have a resistance of 9.666 ohms.
As these three pieces (R1, R2 and R3) are now connected in parallel, the equivalent resistance R can be calculated using this equation:
1/R = 1/R1 + 1/R2 + 1/R3
1/R = 1/9.666 + 1/9.666 + 1/9.666
1/R = 3/9.666
R = 9.666/3 = 3.222 ohms
The resistance between A and B will be 3.222 ohms
Answer:
it is sublimation because There are three ways heat is transferred into and through the atmosphere:
radiation.
conduction.
convection.
Explanation:
please mark me as brainliest as you wrote over there
Answer:
(a) 21.44 ft/s
(b) 0 ft/s
(c) 19.51 ft/s
Explanation:
2 in = 2/12 ft = 0.167 ft
For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:
![v(r) = v_c\left[1 - \frac{r^2}{R^2}\right]](https://tex.z-dn.net/?f=v%28r%29%20%3D%20v_c%5Cleft%5B1%20-%20%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%5Cright%5D)
where R = 0.167 ft is the pipe radius and 
is the constant fluid velocity at the center of the pipe.
We can integrate this over the cross-section area of the in order to find the volume flow
![\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%20%3D%20%5Cint%5Climits%20%7Bv%28r%29%7D%20%5C%2C%20dA%20%5C%5C%3D%20%5Cint%5Climits%5ER_0%20%7Bv_c%5Cleft%5B1%20-%20%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%5Cright%5D2%5Cpi%20r%7D%20%5C%2C%20dr%5C%5C%20%20%3D%202%5Cpi%20v_c%5Cint%5Climits%5ER_0%20%7Br%20-%20%5Cfrac%7Br%5E3%7D%7BR%5E2%7D%7D%20%5C%2C%20dr%5C%5C%20%3D%202%5Cpi%20v_c%20%5Cleft%5B%5Cfrac%7Br%5E2%7D%7B2%7D%20-%20%5Cfrac%7Br%5E4%7D%7B4R%5E2%7D%5Cright%5D%5ER_0%5C%5C%3D%202%5Cpi%20v_c%20%5Cleft%28%5Cfrac%7BR%5E2%7D%7B2%7D%20-%20%5Cfrac%7BR%5E4%7D%7B4R%5E2%7D%5Cright%29%5C%5C%3D%202%5Cpi%20v_c%20%5Cleft%28%5Cfrac%7BR%5E2%7D%7B2%7D%20-%20%5Cfrac%7BR%5E2%7D%7B4%7D%5Cright%29%5C%5C%3D%202%5Cpi%20v_c%20R%5E2%2F4%5C%5C%3D%5Cpi%20v_c%20R%5E2%2F2%5C%5CA%20%3D%20%5Cpi%20R%5E2%5C%5C%5Cdot%7BV%7D%20%3D%20Av_c%2F2%5C%5C)
So the average velocity
b) At the wall of the pipe, r = R so 
c) At a distance of 0.6 in = 0.6/12 = 0.05 ft
Gap junctions in the intercalated discs allow impulses to be spread across the heart more quickly. This is because gap junctions allow particles/signals to pass through, thus making cells with gap junctions more able to interact.
One more thing—you posted this in the physics section rather than biology.
