Answer:
Maximum amount of heat = 10002151.38J
Explanation:
Workdone by motor in 86.1 minutes I given by:
W = power × time
W= 294 × 86.1×60
W= 1439424 Joules
W= 1.4 ×10^6Joules
The amount of heat extracted is given by:
/QL /= K/W/ = TL/W/ /(TH - TL)
Where TL= freezing compartment temperature
TH = Outside air temperature
/QL /= 271 × 1439424 / (310 - 271)
/QL/ = 390083904/39
/QL/ = 10002151.38 Joules
Answer:
confounding cause they had exposure to many programmes
It evaporates into a vapor
~~~hope this helps~~~
~~have a beautiful day~~
~davatar~
Answer:
1. t = 0.0819s
2. W = 0.25N
3. n = 36
4. y(x , t)= Acos[172x + 2730t]
Explanation:
1) The given equation is

The relationship between velocity and propagation constant is

v = 15.87m/s
Time taken, 

t = 0.0819s
2)
The velocity of transverse wave is given by


mass of string is calculated thus
mg = 0.0125N

m = 0.00128kg


0.25N
3)
The propagation constant k is

hence

0.036 m
No of wavelengths, n is

n = 36
4)
The equation of wave travelling down the string is
![y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%3DAcos%5Bkx%20-wt%5D%5C%5C%5C%5Cbecomes%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B%28172%20rad.m%29x%20%2B%20%282730%20rad.s%29t%5D)
![without, unit\\\\y(x , t)= Acos[172x + 2730t]](https://tex.z-dn.net/?f=without%2C%20unit%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B172x%20%2B%202730t%5D)