Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
So, the force between the charges q and Q - q is given by
For maxima and minima, differentiate the force with respect to q.
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now 
the double derivate is negative, so the force is maxima when q = Q / 2 .
Answer:
a) 0.018 kg
b) 262 kPa
Explanation:
The volume of the concentric cylinders would be:
V = π/4 * h * (D^2 - d^2)
V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3
The state equation of gases:
p * V = m * R * T
Rearranging:
m = (p * V) / (R * T)
R is 287 J/(kg * K) for air
25 C = 298 K
m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg
After pumping more air the volume remains about the same, but temperature and pressure change.
30 C = 303 K
m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg
The mass that was added is
m1 - m0 = 0.057 - 0.039 = 0.018 kg
If that air is cooled to 0 C
0 C is 273 K
p = m * R * T / V
p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa
First we need to convert the mm to inches to make our computation
easier.
1mm = 0.0393701
32mm * 0.0393701 = 1.25 in
Solution:
C = 1/2d = ½ (1.25) = 0.625 in^4
Tension: tension = Te/J = 2T/ piC^3
= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi
Bending:
I = pi/4 * c^4 = 119.842 x 10^-3 in^4
M = (5)(600) = 3600 lb in
G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2
psi = -18.775ksi
Gx = -18.775 ksi
Gy = 0
Txy = 6.519 ksi
G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi
1.
G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi
G2 = Gave - R = -9/387 - 11.429 = -20.8
Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 =
-1.3889
ϴp = -27.1 degrees and 62.9 degrees
2.
Tmax = R = 11.43 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2
+ (6.519)^2 = 11.429 ksi
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Answer small rual villages
<span>Increased motion increases pressure. This is why warmer (more energy, moving faster) gases have more pressure than colder. They bump into each other more. Check your tire pressure now, then drive down the road a bit. Check again.
hopefully that helped :)</span>
