Option C
The fact that voltage can be created by exerting force on a crystal is used in Knock sensor
<u>Explanation:</u>
Any knock to an engine exhibits as a little shake that is distinguished by the knock sensor. This sensor acts by altering the fluctuation to an electrical sign, which is later transferred to the processor mastering the ignition system.
There the variation in quake to the voltage sign modifies the timing improvements on the kindling. The knock sensor is placed on the engine base, cylinder cap or consumption manifold. This is because its purpose is to sense fluctuations affected by engine knock or explosion.
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325



as the volume remains constant therefore



therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 

Final mass 

Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
The period of the wave would be halved
Answer ) Sound level equation
The intensity of a sound wave is related to its amplitude squared by the following relationship: I=(Δp)22ρvw I = ( Δ p ) 2 2 ρ v w . Here Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2.
If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.