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Answer:
47947.52 J.
Explanation:
From the question,
Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1
Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.
Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.
Substitute these values into equation 1 above
Q = 640×4.816(48)
Q = 147947.52 J.
Hence the amount of heat given off is 47947.52 J.
Answer:
I₃ /I₀ = 25%
Explanation:
For this exercise let's use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
When the polarized light passes the first polarizer, the intensity is constant if it is in the direction of polarization.
I₁ = I₀
the light transmitted by the second polarizer is
I₂ = I₁ cos² θ
I₂ = I₀ cos² 45
I₂ = I₀ 0.5
this light is polarized 45º with respect to the first polarizer
Now let's examine the third polarizer, it is indicated that it has an angle of 90º with respect to the first polarizer, the angle with respect to the second polarizer is
θ = 90 - 45
θ = 45º
so the polarizer is at an angle of 45º
let's use Maus's law
I₃ = I₂ cos² 45
I₃ = I₀ 0.5 cos² 45
I₃ = I₀ 0.25
the light that passes through the system is
I₃ /I₀ = 0.25
as a percentage we multiply by 100
I₃ /I₀ = 25%
Answer: image to much to type.
Explanation: