Answer:
C
Explanation:
To melt the alcohol
Heat needed = M . L = 2 . 25 = 50 kcal
To warm up the alcohol
Heat needed = M . sp. ht. . ∆t = 2 . 0.6 . 100 = 120 kcal
Total heat needed = 170 kcal
Assuming that 0.6 kcal/ kg / ˚C is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)
Answer:
hey did you complete the whole test?
Explanation:
I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B
Answer:
![\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%203.6%20%5C%20ns%2F%20km%7D)
Explanation:
From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.
Hence, the material dispersion is ![\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )](https://tex.z-dn.net/?f=%5Cdfrac%20%7Bd%20%5Ctau%20_%7Bmat%7D%7D%7Bd%20%5Clambda%20%7D%20%5Csimeq%20%2880%20%5C%20ps%20%2F%20%28nm.km%29%20%5C%20%29)
Now, using the pulse spread formula:
![\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%20%5Cdfrac%7Bd%20%5Ctau%20_%7Bmat%7D%20%7D%7Bd%20%5Clambda%7D%20%5Csigma%20%5Clambda)
![\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%20%2880%20%5C%20ps%2F%20%28%20m.km%29%20%5C%20%29%20%20%5Ctimes%20%2845%20%5C%20nm%29)
Thus, the pulse spreading as a result of material dispersion is:![\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%203.6%20%5C%20ns%2F%20km%7D)
Answer:
a is the answer is yes I....
Explanation:
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