Answer:
fundamental frequency of pipe will be equal to 74 Hz
Explanation:
We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz
Speed of the sound in air is 343 m/sec
We have to find the fundamental frequency for the pipe
Fundamental frequency will be equal to difference of the two adjacent frequency
So fundamental frequency = 370 - 296 = 74 Hz
So fundamental frequency of pipe will be equal to 74 Hz
Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:

where q is the charge and

is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is

, so we can find the charge transferred by the battery:
Answer:
Work = power * time
time = 20000 joules / 1000 joules / sec = 20 sec
Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans