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anastassius [24]
4 years ago
12

After Hurricane Katrina there was considerable public outrage that many of the properties were not insured against flooding alth

ough they were insured against wind damage. What might explain these different approaches to​ insurance? A. Predatory insurance policies. B. The risk of flood damage is potentially diversifiable but the risk of wind damage is not. C. The risk of wind damage is potentially diversifiable but the risk of flooding is not. D. Neither the risk of wind damage or the risk of flooding is diversifiable.
Engineering
2 answers:
Mashcka [7]4 years ago
8 0
I would say B is the answer
Anettt [7]4 years ago
6 0
Indeed, B is the answer
You might be interested in
A 600-ha farmland receives annual rainfall of 2500 mm. There is a river flowing through the farmland with an inflow rate of 5 m3
shepuryov [24]

Answer:

E = 7333.33 mm

Explanation:

The annual evapotranspiration (E) amount can be calculated using the water budget equation:

P*A + Q_{in}*\Delta t = E*A + \Delta S + Q_{out}*\Delta t   (1)

<u>Where</u>:

<em>P: is the precipitation = 2500 mm, </em>

<em>Q(in): is the water flow into the river of the farmland = 5 m³/s, </em>

<em>ΔS: is the change in water storage = 2.5x10⁶ m³,  </em>

<em>Q(out): is the water flow out of the river of the farmland = 4 m³/s.</em>

<em>Δt: is the time interval = 1 year = 3.15x10⁷ s </em>

<em>A: is the surface area of the farmland = 6.0x10⁶ m² </em>  

Solving equation (1) for ET we have:

E = \frac{P*A + Q_{in}*\Delta t - \Delta S - Q_{out}*\Delta t}{A}

E = \frac{2.5 m \cdot 6.0 \cdot 10^{6} m^{2} + 5 m^{3}/s \cdot 3.15 \cdot 10^{7} s - 2.5 \cdot 10^{6} m^{3} - 4 m^{3}/s \cdot 3.15 \cdot 10^{7} s}{6.0\cdot 10^{6} m^{2}}                                  

E = 7333.33 mm

Therefore, the annual evapotranspiration amount is 7333.33 mm.

I hope it helps you!  

3 0
4 years ago
The four important principles of flight are lift, drag, thrust, and _____.
lbvjy [14]

Answer:

gravity force or weight

6 0
4 years ago
A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom
swat32

Answer:

1. Volume of the glass shell (Vg) is simply volume of the empty part of the jar (Ve) subtracted from volume of the entire jar (Vj):

Vg = Vj - Ve

Volume is calculated as base (B) multiplied with height (h). Base of the jar is circle, so its surface is πr^2 (r being the radius).

However radius is different depending on the part of the jar; for empty part of the jar, inner radius is d = 3 in, for the whole jar it is inner radius plus thickness of the glass a = 3 + 3/16 = 3.1875 in.

We are also given height of the whole jar, h = 6 in, but height of the empty part is entire height minus thickness of the jar h' = 6 - 0.1875 = 5.8125 in.

Now, let's calculate:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

So, volume of the glass shell is Vj - Ve which is 27.16 in^3.

2. Mass of the glass jar is density of the glass multiplied with volume:

m = ρ • Vg

Density of the glass is given here in cubic feet so, first, we need to convert it to cubic inches, dividing it by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

So, mass of the jar is:

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To find weight and volume of the water displaced we first need to find how deep the jar sinks (H), because volume of the displaced water is equal to the volume of the jar submerged. Jar will sink until gravity force (pulling it down) and buoyancy force (pushing it up) become equal. Displaced water is πa^2 • H and the buoyancy is ρw • g • Vd (ρw is density of water which is 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is displaced water).

So, buoyancy is:

B = ρw • g • πa^2 • H

We said that buoyancy must be equal to gravity:

B = m • g (m being mass of the jar). So:

ρw • g πa^2 • H = m • g

ρw • πa^2 • H = m

From this, we can find H:

H = m / ρw•πa^2

H = 2.25 inches

That means that the jar will sink 2.25 inches in the water.

3. Now, it's easy to find volume of displaced water. It's the same as the volume of the jar submerged:

Vd = πa^2 • H

Vd = 71.94 in^3

4. And finally, the weight of water is:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

Of course, we see that the mass of the jar equals the mass of the displaced water. Taking this as a rule, this question could have been solved easier However I wanted to do it more detailed, to explain it more clearly

6 0
3 years ago
The input power for a thermostat is wired to the
masya89 [10]

Answer:

A. R

Explanation:

There are basically two wires that supply input power to the thermostat namely the C wire which is the common wire and the R wire .The G wire simply completes the input power circuit from the R-leg of the power supply.On the another hand the Y wire also completes the circuit  from the compressor fan contactor to the R-leg of the power supply . While the W wire completes the path to the heater contactor coil.

6 0
4 years ago
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be
hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0
3 years ago
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