She remains the only person to be honored for accomplishments in two separate sciences. Curie received the Nobel Prize in Physics in 1903, <u>along with her husband and Henri Becquerel, </u>for their work on radioactivity.
<h3>What did Marie Curie discover?</h3>
Relentless regardless of a vocation of truly requesting and at last lethal work, she found polonium and radium, supported the utilization of radiation in medication and essentially changed how we might interpret radioactivity. Curie was conceived Marya Skłodowska in 1867 in Warsaw.
- Curie was the first person to win two Nobel Prizes.
- She managed it all without a fancy lab.
- Nobel Prizes were a family affair.
- Curie was the first female professor at Sorbonne University.
- Curie is buried in the Panthéon in Paris.
To learn more about Marie Curie from the given link
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Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m
An absorption spectrum occurs when light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.
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Yp(t) = A1 t^2 + A0 t + B0 t e(4t)
=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]
y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)
=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)
y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)
Now substitute the values of y ' and y '' in the differential equation:
<span>y′′+αy′+βy=t+e^(4t)
</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)
Next, we equate coefficients
1) Constant terms of the left side = constant terms of the right side:
2A1+ 2αA0 = 0 ..... eq (1)
2) Coefficients of e^(4t) on both sides
8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)
3) Coefficients on t
2αA1 + βA0 = 1 .... eq (3)
4) Coefficients on t^2
βA1 = 0 ....eq (4)
given that A1 ≠ 0 => β =0
5) terms on te^(4t)
16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)
Given that B0 ≠ 0 => 16 + 4α + β = 0
Use the value of β = 0 found previously
16 + 4α = 0 => α = - 16 / 4 = - 4.
Answer: α = - 4 and β = 0