A3.49 rad/s (33 rpm) record has a 3.98-khz tone cut in the groove. if the groove is located 0.100 m from the center of the record (see drawing), what is the "wavelength" in the groove?

8.77×10⁻⁵ m

Explanation:

ω = angular velocity = 3.49 rad/s = 33 rpm

r = Radius of the groove = 0.1 m

f = Frequency = 3.98 kHz

Tangential Velocity of record

v = ωr

⇒v = 3.49×0.1

⇒v = 0.349 m/s

The velocity of the wave is 0.349 m/s

v = fλ

∴ Wavelength in the groove is 8.77×10⁻⁵ m

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas