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2 years ago

A properly fitted wearable pfd should have which characteristics

Engineering

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

Answer:a composit

Explanation:

it is literally called it

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A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int

Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

$s = v_0t + \frac{at^2}{2}$

$s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)$

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0

3 years ago

A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and

vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium K = 205 W/m k

heat flux $= k \frac{dt}{dx}$

$= 205 \frac{325 - 275}{0.02}$

= 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 16.3 \frac{325 - 275}{0.02}$

= 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 0.8 \times \frac{325 - 275}{0.02}$

= 2 KW/m2

6 0

3 years ago

A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-c

DanielleElmas [232]

Answer:

Explanation:

5 0

3 years ago

Technician A says when tightening lug or wheel nuts a torque wrench must be used. Technician B says that an air wrench can be us

ASHA 777 [7]

A, is your answer hope it helps

4 0

3 years ago

If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b

zalisa [80]

Answer:

h=32.1 km

Explanation:

<em>solution:</em>

using newton law of gravitational attraction and newton second law:

$W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\$

$m_{E}= mass of earth$

r= distance between two masses

at sea level

a=g

$r=R_{E}$

$a=\frac{Gm_{E}}{r^{2}}$ .............................(1)

$Gm_{E} =gR_{E}^2$ .........................(2)

by substituting (2) and (1) $a=g\frac{R_{E}^2 }{r^{2} }$ acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0

3 years ago