Answer:
Map and avoid high-risk zones.
Build hazard-resistant structures and houses.
Protect and develop hazard buffers (forests, reefs, etc.)
Develop culture of prevention and resilience.
Improve early warning and response systems.
Build institutions, and development policies and plans.
Explanation:
Answer:
The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Explanation:
Given that,
Thickness of A = 8.0 m
Conductivity = 25.0 m/d
Thickness of B = 2.0 m
Conductivity = 142 m/d
Thickness of C = 34 m
Conductivity = 40 m/d
We need to calculate the horizontal conductivity
Using formula of horizontal conductivity
![K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}](https://tex.z-dn.net/?f=K_%7BH%7D%3D%5Cdfrac%7BH_%7BA%7DK_%7BA%7D%2BH_%7BA%7DK_%7BA%7D%2BH_%7BA%7DK_%7BA%7D%7D%7BH_%7BA%7D%2BH_%7BB%7D%2BH_%7BC%7D%7D)
Put the value into the formula
![K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}](https://tex.z-dn.net/?f=K_%7BH%7D%3D%5Cdfrac%7B8.0%5Ctimes25%2B2%2C0%5Ctimes142%2B34%5Ctimes40%7D%7B8.0%2B2.0%2B34%7D)
![K_{H}=41.9\ m/d](https://tex.z-dn.net/?f=K_%7BH%7D%3D41.9%5C%20m%2Fd)
We need to calculate the vertical conductivity
Using formula of vertical conductivity
![K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}](https://tex.z-dn.net/?f=K_%7BV%7D%3D%5Cdfrac%7BH_%7BA%7D%2BH_%7BB%7D%2BH_%7BC%7D%7D%7B%5Cdfrac%7BH_%7BA%7D%7D%7BK_%7BA%7D%7D%2B%5Cdfrac%7BH_%7BB%7D%7D%7BK_%7BB%7D%7D%2B%5Cdfrac%7BH_%7BC%7D%7D%7BK_%7BC%7D%7D%7D)
Put the value into the formula
![K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}](https://tex.z-dn.net/?f=K_%7BV%7D%3D%5Cdfrac%7B8.0%2B2.0%2B34%7D%7B%5Cdfrac%7B8.0%7D%7B25%7D%2B%5Cdfrac%7B2.0%7D%7B142%7D%2B%5Cdfrac%7B34%7D%7B40%7D%7D)
![K_{V}=37.2\ m/d](https://tex.z-dn.net/?f=K_%7BV%7D%3D37.2%5C%20m%2Fd)
Hence, The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Answer:
hello your question is incomplete attached below is the complete question
answer : attached below
Explanation:
let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200![\pi](https://tex.z-dn.net/?f=%5Cpi)
g(t) = x ( t ) sin ( 2000 ![\pi t )](https://tex.z-dn.net/?f=%5Cpi%20t%20%29)
![x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )](https://tex.z-dn.net/?f=x_%7B1%7D%20%28t%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20x%28t%29%20%20sin%20%28%204000%5Cpi%20t%20%29)
next we apply Fourier transform
attached below is the remaining part of the solution
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa, ![V_{1} = 15 m^{3}/s](https://tex.z-dn.net/?f=V_%7B1%7D%20%3D%2015%20m%5E%7B3%7D%2Fs)
![T_{1} = 27^{o}C = (27 + 273) K = 300 K](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%2027%5E%7Bo%7DC%20%3D%20%2827%20%2B%20273%29%20K%20%3D%20300%20K)
We know that for an ideal gas the mass flow rate will be calculated as follows.
![P_{1}V_{1} = mRT_{1}](https://tex.z-dn.net/?f=P_%7B1%7DV_%7B1%7D%20%3D%20mRT_%7B1%7D)
or, m = ![\frac{P_{1}V_{1}}{RT_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BRT_%7B1%7D%7D)
=
= 10 kg/s
Now, according to the steady flow energy equation:
![mh_{1} + Q = mh_{2} + W](https://tex.z-dn.net/?f=mh_%7B1%7D%20%2B%20Q%20%3D%20mh_%7B2%7D%20%2B%20W)
![h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}](https://tex.z-dn.net/?f=h_%7B1%7D%20%2B%20%5Cfrac%7BQ%7D%7Bm%7D%20%3D%20h_%7B2%7D%20%2B%20%5Cfrac%7BW%7D%7Bm%7D)
![C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}](https://tex.z-dn.net/?f=C_%7Bp%7DT_%7B1%7D%20-%20%5Cfrac%7B80%7D%7B10%7D%20%3D%20C_%7Bp%7DT_%7B2%7D%20-%20%5Cfrac%7B130%7D%7B10%7D)
![(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}](https://tex.z-dn.net/?f=%28T_%7B2%7D%20-%20T_%7B1%7D%29C_%7Bp%7D%20%3D%20%5Cfrac%7B130%20-%2080%7D%7B10%7D)
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
= ![32^{o}C](https://tex.z-dn.net/?f=32%5E%7Bo%7DC)
Therefore, we can conclude that the exit temperature of the gas in deg C is
.