A person riding in a hot air air balloon basket ascends to 1000 ft above the ground. What happened to the atmospheric pressure a
1 answer:
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Answer:
a)
b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c) The material is peat, possibly.
d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Explanation:
Given:
- mass of aluminium,
- mass of water,
- initial temperature of the system,
- mass of copper block,
- temperature of copper block,
- mass of the other block,
- temperature of the other block,
- final equilibrium temperature,
We have,
specific heat of aluminium,
specific heat of copper,
specific heat of water,
Using the heat energy conservation equation.
The heat absorbed by the system of the calorie-meter to reach the final temperature.
The heat released by the blocks when dipped into water:
where
specific heat of the unknown material
For the conservation of energy :
so,
b)
This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c)
The material is peat, possibly.
d)
The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º