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3 years ago

Which of the following conditions were present in over 80% of paddling fatalities from 1995-2000?

Engineering

1 answer:

Minchanka [31]3 years ago

3 0

Answer:

80% of the people that were killed weren't wearing a safety flotation device ( in correct terminology Personal Flotation Device, or PFD )

Explanation:

Hence they drowned due to the lack of safety.

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The interactive activities in this course

MariettaO [177]

Answer:

Explanation:

3 0

3 years ago

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where

Harrizon [31]

Answer:

a. 430.944 pascal

b. 0.0625psi

c. 1.73008inH20

Explanation:

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where p is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P-1.2 kg/m and V-60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column

checking the dimensional consistency

Dp= $\frac{kg}{m^3} (\frac{m^2}{s^2} \\\frac{x}{y} \\\\\frac{kg}{ms^2} \\which s the same as\\ \frac{N}{m^2}$

$Dp=∈(\frac{v^2}{2} )$

convert 1 mile to meter

1mile=1609m

1h=3600s

60mile/h=26.8m/s

slotting intpo the relation

$Dp=1.2kg/m^3(\frac{26.8^2}{2} )$

430.944kg/(ms^2)

which is the same as 430.944N/m^2

expressing in pascal.

We know that

1 pascal=1 N/m^2

430.944 pascal

2. 1 pascal=0.000145psi

answer=0.0625psi

3.1 pascal=0.00401inH20

answer=430.944*0.0040146

1.73008inH20

5 0

3 years ago

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in t

uysha [10]

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

8 0

3 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

alisha [4.7K]

Answer:

The level of the service is loss and the density is 34.2248 pc/mi/ln

Explanation:

the solution is attached in the Word file

Download docx

6 0

3 years ago

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

3 years ago