First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R = ![200\mu m = 200\times 10^{- 6}\ m](https://tex.z-dn.net/?f=200%5Cmu%20m%20%3D%20200%5Ctimes%2010%5E%7B-%206%7D%5C%20m)
Gauge Pressure inside, ![P_{in} = 25\ mmHg](https://tex.z-dn.net/?f=P_%7Bin%7D%20%3D%2025%5C%20mmHg)
Blood Pressure outside, ![P_{o} = 10\ mmHg](https://tex.z-dn.net/?f=P_%7Bo%7D%20%3D%2010%5C%20mmHg)
Now,
Change in pressure, ![\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%2025%20-%2010%20%3D%2015%5C%20mmHg%20%3D%201.99%5Ctimes%2010%5E%7B3%7D%5C%20Pa)
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:
![\Delta P = \frac{4\pi T}{R}](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B4%5Cpi%20T%7D%7BR%7D)
![T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B1.99%5Ctimes%2010%5E%7B3%7D%5Ctimes%20200%5Ctimes%2010%5E%7B-%206%7D%7D%7B4%5Cpi%7D%20%3D%200.0318%5C%20N%2Fm%20%3D%200.318%5C%20mN%2Fm)
And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
I think 36m/12s because 3×12 =36