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tekilochka [14]
4 years ago
12

Why was the lack of preparedness of the Federal Emergency Management Agency in the Hurricane Katrina disaster so damaging for th

e George W. Bush administration? a. President Bush had vociferously denied that hurricanes of this size could ever reach the United States. b. The Bush administration had prided itself on its unique focus for homeland security. c. There was nothing unusual about this particular hurricane, but the city and state had not done anything to prepare the city’s population. d. As a native of Louisiana, he was expected to display particular care for New Orleans. e. President Bush had been the head of FEMA during his father’s presidency.
Engineering
1 answer:
kolbaska11 [484]4 years ago
8 0

Answer:

B

Explanation:

The Bush administration had prided itself on its unique focus for homeland security and combating terrorism, without adequate preparation for the disaster.

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Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?
WITCHER [35]

Answer:

You can create high drag which allows a steeper angle without increasing your air speed on landing. you can reduce the length of landing role. Flaps are also used to increase the drag they are retracted when they are not needed. it is adviseable to down he flaps during the time of take off.

4 0
3 years ago
Basic Question please help
Dvinal [7]
1(A)
2(B)
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7 0
3 years ago
In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images
Zarrin [17]

Answer:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

A= \frac{r^5 \rho}{t^2}

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant \rho_{air}.

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= \frac{ML^2}{T^2} A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as \frac{L}{T}

[t]=T represent the time

[\rho]=\frac{M}{L^3} represent the density.

Solution to the problem 

And if we analyze the function for r we got this:

[r]=L=[A]^x [\rho]^y [t]^z

And if we replpace the formulas for each on we got:

[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z

And using algebra properties we can express this like that:

[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

x+y =0 , 2x-3y=1, -2x+z=0

We can solve for x like this x=-y and replacing into quation 2 we got:

2(-y)-3y = 1

-5y = 1

y= -\frac{1}{5}

And then we can solve for x and we got:

x = -y = -(-\frac{1}{5})=\frac{1}{5}

And if we solve for z we got:

z=2x =2 \frac{1}{5}=\frac{2}{5}

And now we can express the radius in terms of the dimensional analysis like this:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}

A= \frac{r^5 \rho}{t^2}

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed \rho =1.2 kg/m^2, and replacing we got:

A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}

And we can convert this into ergs we got:

A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs

And then we know that 1 g of TNT have 4x10^4 erg

And we got:

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

3 0
3 years ago
How is the angle between power strokes determined?
aivan3 [116]
And there's an equation to help determine which configurations will work best. In a four-stroke engine, an individual piston fires every 720 degrees (two crankshaft rotations). ... The flat-four fires at 180-degree intervals, and its V angle is 180 degrees, which leads to a balance of firing forces.Jan 14, 2011

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5 0
3 years ago
A hypothetical metal alloy has a grain diameter of 1.7 ´ 10-2 mm. After a heat treatment at 450°C for 250 min the grain diameter
Yuliya22 [10]

Answer:1103 minutes/66180 seconds

Explanation:

FIRST STEP: is to FIND the value of K using the formula below;

K= d^n - d^n(o)/t .....................(1).

Parameters given from the question d^n = 4.5×10^-2 mm, d^n(o)= 1.7 × 10^-2 mm, t= 250 minutes(min) and n= 2.1.

Slotting in the parameters into the equation (1) above,then;

(4.5×10^-2)^2.1 - (1.7×10^-2)^2.1/ 250

= 5.2 × 10^-6 mm^2.1/min.

SECOND STEP: with the value of K from the second step, we can use it to calculate the required time based on the diameter. Therefore, equation (1) becomes;

t= d^2.1 - d^2.1(o)/ K

(8.7×10^-2)^2.1 - (1.7×10^-2)^2.1/ 5.2 × 10^-6

= 1,103 minutes.

5 0
4 years ago
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