1. The chemical reaction produced by Carlo's fire is exergonic because energy is "going out". As the reaction proceeds, entropy increases as the energy stored in the dry wood and leaves are used up as fuel to create the fire which produces low quality light and warmth.
2. This reaction is a classic example of an exothermic reaction. Exothermic reactions are characterized with the presence of heat and light in the products. Combustion reactions are always exothermic in nature.
3. Catalyst are substances that are used to speed up reactions by lowering the activation requirement. Catalysts aren't consumed in the reaction and can still be chemically retrieved afterwards. In this situation, the leaves cannot be retrieved after the reaction ends. The leaves speed up the heating of the wood but it does not behave as a catalyst.
The Beams And Joints That Hold It .
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
Answer:
200 N = 200 Newtons
Explanation:
Just use the formula F = m*a
F = Force in Newtons
m = mass and is 20 kg
a = acceleration and is 10 m/s^2
F = 20 * 10
F = 200 Newtons.
When an object absorbs an amount of energy equal to Q, its temperature raises by
following the formula
where m is the mass of the object and
is the specific heat capacity of the material.
In our problem, we have
,
and
, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
