Answer:
The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel
" applies to electric charges.
Explanation:
There are only two types of electric charges. Both having own magnitude but different charge.
1. Positive charge
2. Negative charge
Like charges repel each other and opposite charges always attract each other.
When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.
Answer: The final temperature is 470K
Explanation: Using the relation;
Q= ΔU +W
Given, n = 2mol
Initial temperature T1= 345K
Heat =Q= 2250J
Workdone=W=-870J(work is done on gas)
T2 =Final temperature =?
ΔU =3/2nR(T2-T1)
ΔU=3/2 × 2 ×8.314 (T2 - 345)
ΔU=24.942(T2-345)
Therefore Q = 24.942(T2-345)+ (-870)
2250=24.942(T2-345)+ (-870)
125.09=(T2-345)
T2 =470K
Therfore the final temperature is 470K
Answer:
v = 2,99913 10⁸ m / s
Explanation:
The velocity of propagation of a wave is
v = λ f
in the case of an electromagnetic wave in a vacuum the speed that speed of light
v = c
When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium
λ = λ₀ / n
where n is the index of refraction of the material medium.
Therefore the expression is
v =
Let's look for the frequency of blue light in a vacuum
f =
f =
f = 6.667 10¹⁴ Hz
the refractive index of air is tabulated
n = 1,00029
let's calculate
v =
450 10-9 / 1,00029 6,667 1014
v = 2,99913 10⁸ m / s
we can see that the decrease in speed is very small