Answer:
T=575.16K
Explanation:
To solve the problem we proceed to use the 1 law of diffusion of flow,
Here,
is the rate in concentration
is the rate in thickness
D is the diffusion coefficient, where,
Replacing D in the first law,
clearing T,
Replacing our values
The speed of light and the propagation of errors allows to find the results on the questions of the radiation emitted by the laser are:
a) The frequency is: f = 3.7 10¹⁴ Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
a) The speed of a wave is related to its wavelength and frequency.
c = λ f
Where c is the speed of light, λ the wavelength and f the frequency.
They indicate that the wavelength is λ = 800 nm = 800 10⁻⁹ m, the speed of light is a constant c = 2.99 10⁸ m/s.
f =
F = 3.7 10¹⁴ Hz
b) Planck's equation states that the energy is proportional to the frequency of the radiation.
E = h f
Where E is the energy, h the Planck constant and f the frequency.
E = 6.63 10⁻³⁴ 3.7 10¹⁴
E = 2.46467 10⁻¹⁹ J
The uncertainty or error is the fluctuation that a magnitude may have due to the precision in the measurements, when the magnitude is calculated by some formula, the propagation of these uncertainties must be carried out.
Δm = ∑
the expression for energy is:
E =
When the error in the measured magnitude is not explicitly indicated, we assume that the error is in the last digit written, therefore
Δλ = ± 1 nm = ± 1 10⁻⁹ m
We calculate.
ΔE = 3.1 10⁻²² J
the error is given with a significant figure.
ΔE = 3 10⁻²² J = 0.004 10⁻¹⁹ J
The result of the energy is:
E = (2.465 ± 0.004) 10⁻¹⁹ J
In conclusion, using the speed of light and the propagation of errors, we can find the results on the questions of the radiation emitted by the laser are:
a) The frequency is; f = 3.7 1014 Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
Learn more here: brainly.com/question/15557220
The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.
Given:
mass of block 1, m₁ = 40 gm = 40×10⁻³ kg
mass of block 2, m₂ = 20 gm = 20×10⁻³ kg
Applied force, F = 0.03 N
Calculation:
Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:
F = ma
where F is the applied force
m is the total mass of the system
a is the acceleration of block 2 (as it is pulled by horizontal force)
From the above equation we get:
0.03 N = (m₁+m₂) a
⇒ a = (0.03 N) / (m₁+m₂)
⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)
⇒ a = 0.5 m/s²
Now, from the free-body diagram of block 2 as shown in figure 3, we get:
Balancing the forces along the horizontal:
∑Fₓ = 0
∴ T = m₂ a
where T is tension in the string
a is the acceleration of block 2
Applying values in the above equation we get:
T = (20×10⁻³ kg) × (0.5 m/s²)
= 0.01 N
Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.
Learn more about tension here:
<u>brainly.com/question/13397436</u>
#SPJ4
