Answer:The energy of the wave by a factor of 4
Explanation:
Answer:

Explanation:
As we know that amplitude of forced oscillation is given as

here we know that natural frequency of the oscillation is given as

here mass of the object is given as



angular frequency of applied force is given as


now we have


<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at
is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
........(1)
Velocity of sound wave is given by :




Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
( n = integers )
Let n = 1
So, 


Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.
neutron star is formed when the core of a star, that was one ginormous, collapse on its own mass thus it loses it's volume exponentially (y=a^x) while the mass of the star decreases linearly (y=mx+c) thus the neutron star has mass of 10 to 29 sun in the radius of 10 km( for scale earth has 6731 km) thus it's density is super high!!!!!!!!!!!!!!!!!!!
hope it helps you! ;)
Answer:
135,000 m
Explanation:
(1500 m/s) (90 s) = 135,000 m