Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
true
because if rain gets in your engine it could destroy it or you can get electrocuted if you try to touch your engine
Answer:
The initial temperature is 649 K (376 °C).
The final pressure is 0.965 MPa
Explanation:
From the ideal gas equation
PV = nRT
P is the initial pressure of water = 2 MPa = 2×10^6 Pa
V is intial volume = 150 L = 150/1000 = 0.15 m^3
n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol
R is gas constant = 8.314 m^3.Pa/mol.K
T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C
From pressure law,
P1/T1 = P2/T2
P2 (final pressure) = P1T2/T1
T2 (final temperature) = 40 °C = 40 + 273 = 313 K
P1 (initial pressure) = 2 MPa
T1 (initial temperature) = 649 K
P2 = 2 × 313/649 = 0.965 MPa
Answer:
AC voltages alternate/cycle, at a rate of 60 times each second.
Answer:
a) 
b) 
Explanation:
From the properties of Super-heated Refrigerant 134a Vapor at 
, 
; we obtain the following properties for specific enthalpy and specific entropy.
So; specific enthalpy 
specific entropy 
Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at 
; we obtain the following properties:
Given that the power input to the compressor is 2 hp;
Then converting to Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr
2 hp = 2 × 2544.4342 Btu/hr
2 hp = 5088.8684 Btu/hr
The steady state energy for a compressor can be expressed by the formula:
By neglecting kinetic and potential energy effects; we have:
b) To determine the entropy generation; we employ the formula:
In a steady state condition 
Hence;
![\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]](https://tex.z-dn.net/?f=%5Csigma%20_c%20%3D%20%5B200%20%5C%20lb%2Fhr%20%280.2157%20-0.2315%29%20%5C%20Btu%2Flb%20.%5E0R%20%20-%20%5Cdfrac%7B%28-3730.8684%20%5C%20Btu%2Fhr%29%7D%7B%2840%5E0%20%2B%20459.67%5E0%29%5E0R%7D%5D)
![\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}]](https://tex.z-dn.net/?f=%5Csigma%20_c%20%3D%20%5B%28-3.16%20%29%20%5C%20Btu%2Fhr%20.%5E0R%20%20%2B%20%287.4667%20%29%20Btu%2Fhr%20%5E0R%7D%5D)
