Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have
, where:
is the length

We divide the first equation by the second equation to get:


Using this Area, we find the diameter of the wire:



To find the length, we multiply the two equations stated initially:


Answer:
(A) 140 j/sec (b) 1.26 K
Explanation:
We have given the heat heat flowing into the refrigerator = 40 J/sec
Work done = 40 W
(a) So the heat discharged from the refrigerator 
(b) Total heat absorbed =140 j/sec 
Let the temperature be 
Heat absorbed per hour =504000 ![[tex]=400\times 10^3\times \Delta T](https://tex.z-dn.net/?f=%5Btex%5D%3D400%5Ctimes%2010%5E3%5Ctimes%20%5CDelta%20T)
So 
Explanation:
When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017
The Answer is:
O 3s
Hope you got it right.