Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)
Solids' molecules are strongly attracted to one another. As a result, the molecules are barely moving and tightly packed. Because of this, shape and volume are fixed.
The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.
The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).
You can learn more about states of the matter here:
brainly.com/question/18538345
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Answer:
.
Explanation:
Let 
denote the absolute temperature of this object.
Calculate the value of before and after heating:
.
.
By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to 
.
Ratio between the absolute temperature of this object before and after heating:
.
Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:
.
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
Guess I recommend doing that
The answer is C, as there is not increase or decrease in speed during that time frame.
