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anastassius [24]
3 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]3 years ago
8 0
I would say B is the answer
Anettt [7]3 years ago
6 0
Indeed, B is the answer
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Repetitive movements at work can lead to injuries. True or False
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Answer

True

Explanation

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2 years ago
Which option shows the most valuable metallic properties
Rina8888 [55]

Malleable and ductile

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6 0
2 years ago
A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is
Talja [164]

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × \pi × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × \pi × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque =  30316.369 lb-in

3 0
3 years ago
Which option identifies the tool best to use in the following scenario?
Lena [83]

Answer:

an Allen wrench

Explanation:

it is hexagonal

3 0
3 years ago
A site is underlain by a soil that has a unit weight of 118 lb/ft3. From laboratory shear strength tests that closely simulated
Rzqust [24]

Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

Explanation:

Given that;

Weight of soil r = 118 lb/ft³

stress parameter C = 250 lb/ft²

φ total = 29°

depth Z = 12 ft

The shear strength on a horizontal plane at a depth of 12ft

ζ = C + δtanφ

where δ = normal stress

normal stress δ = r × z = 118 × 12 = 1416

so

ζ = C + δtanφ

ζ = 250 + 1416(tan29°)

ζ = 250 + 1416(tan29°)

ζ = 250 + 784.9016

ζ = 1034.9015 lb/ft²

Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

7 0
3 years ago
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