Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
Wave A would have higher amplitude
Hope this helps :D
Answer:
r₂=0.1 m
Explanation:
Given that
r₁= 10 m , β₁ = 20 dB
At r₂ ,β₂= 60 dB
As we know that intensity level of sound given as



10² x 10⁻¹² = I₁
I₁=10⁻¹⁰ W/m²


10⁶ x 10⁻¹² = I₂
I₂ = 10⁻⁶ W/m²
I₁=10⁻¹⁰ W/m²
P = I A
P=Power ,I =Intensity ,A=Area


r₂=0.1 m