To develop this problem it is necessary to apply the concepts related to Sound Intensity.
By definition the intensity is given by the equation

Where,
I = Intensity of Sound
= Intensity of Reference
At this case we have that 15 engines produces 15 times the reference intensity, that is

And the total mutual intensity is 100 dB, so we should




Therefore each one of these engines produce D. 88dB.
Answer:
a) 27.2 V
b)27.2 V
Explanation:
Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C
Radius = r = 0.53×10⁻¹⁰ m
Electric Potential = V = k q/r
k = 9 ×10⁹ N m²/C² = Coulomb's constant.
V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V
b) Potential Energy of the electron = k q × q / r
= [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,
since 1 electron volt = (1.6 × 10⁻¹⁹)joules
= 27.2 eV
Velocity. Velocity is physical vector quantity; both magnitude and direction are needed to define it. Magnitude is this case meaning speed.