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Answer:
Fnet = 12 N
Explanation:
Force on a point charge due to another point charge = kq1q2 / d^2
Force on +32uC = due to + 20uC + due to -60uC
where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2
Net Force =
![= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%20_0%7D%20%5B%5Cfrac%7B32%20%5Ctimes%2010%5E-%5E6%20%5Ctimes60%5Ctimes10%5E-%5E6%7D%7B%2860%2F100%29%5E2%7D-%5Cfrac%7B32%20%5Ctimes%2010%5E-%5E6%20%5Ctimes20%5Ctimes10%5E-%5E6%7D%7B%20%2840%2F100%29%5E2%7D%20%20%5D)
![F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D9%20%5Ctimes10%5E9%5Ctimes%2010%5E-%5E1%5E2%5B%5Cfrac%7B32%5Ctimes60%5Ctimes10%5E4%7D%7B60%5Ctimes60%7D%20-%5Cfrac%7B32%5Ctimes20%5Ctimes10%5E4%7D%7B40%5Ctimes40%7D%20%5D)
![=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N](https://tex.z-dn.net/?f=%3D90%5B32%28%5Cfrac%7B80-60%7D%7B60%5Ctimes%2080%7D%20%29%5D%5C%5C%5C%5C%3D90%5Ctimes32%5Ctimes0.004167%5C%5C%5C%5C%3D12N)
Fnet = 12 N
Explanation:
(a)
The initial vertical velocity is 13 m/s. At the maximum height, the vertical velocity is 0 m/s.
v = at + v₀
0 = (-9.8) t + 13
t ≈ 1.33 s
(b)
Immediately prior to the explosion, the ball is at the maximum height. Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.
v = √(vx² + vy²)
v = √(25² + 0²)
v = 25 m/s
(c)
Momentum is conserved before and after the explosion.
In the x direction:
m vx = ma vax + mb vbx
m (25) = (⅓ m) (0) + (⅔ m) (vbx)
25m = (⅔ m) (vbx)
25 = ⅔ vbx
vbx = 37.5 m/s
And in the y direction:
m vy = ma vay + mb vby
m (0) = (⅓ m) (0) + (⅔ m) (vby)
0 = (⅔ m) (vby)
vby = 0 m/s
Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.
vy = -13 m/s
And the horizontal velocity will stay constant.
vx = 37.5 m/s
The velocity vector is (37.5 i - 13 j) m/s. The magnitude is:
v = √(vx² + vy²)
v = √(37.5² + (-13)²)
v ≈ 39.7 m/s
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density 
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Answer:
An aqueous stagnant layer that overlies the apical membrane and the subepithelial blood flow are potential barriers to the absorption of drugs that readily penetrate the absorbing cell of the epithelium. The apical, basal, and basement membranes are potential barriers to the absorption of less permeable drugs.
