What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days
Answer:
S= 1.40x10⁻⁵mol/L
Explanation:
The Henry's Law is given by the next expression:
(1)
<em>where S: is the solubility or concentration of Ar in water,
: is Henry's law constant and p: is the pressure of the Ar </em>
<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>
Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.
Have a nice day!
Answer:
Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts