Answer:
Option B. Decreases
Explanation:
Coulomb's law states that:
F = Kq₁q₂ / r²
Where:
F => is the force of attraction between two charges
K => is the electrical constant.
q₁ and q₂ => are the two charges
r => is the distance apart.
From the formula:
F = Kq₁q₂ / r²
The force of attraction (F) is inversely proportional to the square of their separating distance (r).
This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.
Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.
Option B gives the right answer to the question.
Answer:
Option C
Explanation:
Answer C is the correct option. water can be written as H₂O, which means that there are 2 Hydrogen atoms for every oxygen atom, therefore it will occupy more space than oxygen and push more. there is also one more possibility, if the splitting takes place in Hoffman's Voltameter then the Hydrogen will be close to the cathode as hydrogen is positive. Otherwise, option C is correct answer. Hope this Helps you!
Answer:
(a). 14.4 lbf/in^2.
(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Explanation:
So, from the question above we are given the following parameters which are going to help us in solving this particular Question;
=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"
=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."
With these knowledge, let us delve right into the solution;
(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].
The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.
(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.
Length of mercury = 27.8 in.
AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Answer:
122.735 behind converging lens ; 2.16
Explanation:
Given tgat:
Object distance, u = 29 cm
Image distance, v =
Focal length, f = - 19 (diverging lens)
Mirror formula :
1/u + 1/v = 1/f
1/29 + 1/v = - 1/19
1/v = - 1/19 - 1/29
1/v = −0.087114
v = −11.47916
v = -11.48
Second lens
Object distance :
u = 11.48 + 11 = 22.48 cm
1/v = 1/19 - 1/22.48
1/v = 0.0081475
v = 1 / 0.0081475
v = 122.735 cm
122.735 behind second lens
Magnification, m
m = m1 * m2
m = - v / u
Lens1 :
m1 = -11.48 / 29 = - 0.3958620
m2 = - 122.735 / 22.48 = - 5.4597419
Hence,
- 0.3958620 * - 5.4597419 = 2.16
Answer:
0.22 m
Explanation:
= Atmospheric pressure = 101325 Pa
= Pressure at the bottom = tex]2 P_{o}[/tex] = 2 (101325) = 202650 Pa
= height of the container = 7.59 m
= depth of the mercury
Pressure at the bottom = Atmospheric pressure + Pressure due to mercury + Pressure due to water
