Answer:
<h3>a.</h3>
- After it has traveled through 1 cm :

- After it has traveled through 2 cm :

<h3>b.</h3>
- After it has traveled through 1 cm :

- After it has traveled through 2 cm :

Explanation:
<h2>
a.</h2>
For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient
the formula is:

where I is the intensity of the beam,
is the incident intensity and x is the length of the material traveled.
For our problem, after travelling 1 cm:




After travelling 2 cm:




<h2>b</h2>
The optical density od is given by:
.
So, after travelling 1 cm:




After travelling 2 cm:




Answer:
1 ohm
Explanation:
First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:
1 / Eq = (1 / r₁) + (1 / r₂)
The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:
Eq = r₁ + r₂
Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:
1/E₁ = 1/2 + 1/2
1/E₁ = 1
E₁ = 1Ω
1/E₂ = 1/2 + 1/2
1/E₂ = 1
E₂ = 1Ω
This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:
E₃ = E₁ + E₂ = 1 + 1 = 2Ω
The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:
1/Eq = 1/E₃ + 1/2
1/Eq = 1/2 + 1/2
1/Eq = 1
Eq = 1Ω
Yes, ionic bond is formed by the transfer of electrons from and an atom of metallic element to the atom of an non metallic element
Answer:
Answer below!!
Explanation:
In short,the Big Bang hypothesis states that all of the current and past matter in the Universe came into existence at the same time,roughly 13.8 billion years ago.At this time,all matter was compacted into a very small ball with infinite density and intense heat called a Singularity.
Hope I Helped!!!
:)
P=m/v
p=(292.2-130.2)/200
p=0.8100g/ml