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!
You times the 6 by the 350 duvided 1.8
A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.
Answer:
Proof in explanataion
Explanation:
The basic dimensions are as follows:
MASS = M
LENGTH = L
TIME = T
i)
Given equation is:
where,
H = height (meters)
u = speed (m/s)
g = acceleration due to gravity (m/s²)
Sin Ф = constant (no unit)
So there dimensions will be:
H = [L]
u = [LT⁻¹]
g = [LT⁻²]
Sin Ф = no dimension
Therefore,
![[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]](https://tex.z-dn.net/?f=%5BL%5D%20%3D%20%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%5E2%7D%7B%5BLT%5E%7B-2%7D%5D%7D%5C%5C%5C%5C%5C%20%5BL%5D%20%3D%20%5BL%5E%7B%282-1%29%7DT%5E%7B%28-2%2B2%29%7D%5D)
<u>[L] = [L]</u>
Hence, the equation is proven to be homogenous.
ii)
where,
F = Force = Newton = kg.m/s² = [MLT⁻²]
G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²
G = [M⁻¹L³T⁻²]
m₁ = m₂ = mass = kg = [M]
r = distance = m = [L]
Therefore,
![[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\](https://tex.z-dn.net/?f=%5BMLT%5E%7B-2%7D%5D%20%3D%20%5Cfrac%7B%5BM%5E%7B-1%7DL%5E%7B3%7DT%5E%7B-2%7D%5D%5BM%5D%5BM%5D%7D%7B%5BL%5D%5E2%7D%5C%5C%5C%5C%5C%20%5BMLT%5E%7B-2%7D%5D%20%3D%20%5BM%5E%7B%28-1%2B1%2B1%29%7DL%5E%7B%283-2%29%7DT%5E%7B-2%7D%5D%5C%5C%5C%5C)
<u>[MLT⁻²] = [MLT⁻²]</u>
Hence, the equation is proven to be homogenous.
In physical chemistry or in thermodynamics, the work done on the system or by the system (depending on the sign convention) can be determined in several ways. When assumptions like ideal gas behavior is applied, then the formula for work is
W = Δ(PV)
which is the change of the product of Pressure and Volume. But since it was specified that Pressure is constant, the work could be simplified into
W = PΔV = P(V₂ - V₁)
Since we already know the constant pressure and the volumes of the ideal gas before and after the change, we could now solve for work. But let's establish first the units of work which is in Joules. When simplified, Joules is equal to m³*Pa. So, we first change the unit of pressure from atm to Pascals ( 1 atm = 101,325 Pa) and the unit of volume from liters to m³ (1 m³ = 1000 L),
1.5 atm * 101325 Pa/1 atm = 151987.5 Pa
15 L * 1 m³/1000 L = 0.015 m³
35 L * 1 m³/1000 L = 0.035 m³
Then, they are now ready for substitution,
W = 151987.5 Pa (0.035 m³ - 0.015 m³)
W = 3,039.75 Joules
