Answer:
Sohan told Geeta that i had done my work
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Answer: Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.
Explanation:
I think D. liquid water moving along the surfac
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
