<h2>
Answer: 0.17</h2>
Explanation:
The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":
(1)
Where:
is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 
is the Stefan-Boltzmann's constant.
is the Surface area of the body
is the effective temperature of the body (its surface absolute temperature) in Kelvin.
However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:
(2)
Where 
is the body's emissivity
(the value we want to find)
Isolating from (2):
(3)
Solving:
(4)
Finally:
(5) This is the body's emissivity
Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m
Answer:
yes
it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
