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3 years ago

All sized companies are required to have a written fire prevention plan true or false

Engineering

2 answers:

zmey [24]3 years ago

7 0

Answer:

The statement is true. All sized companies are required to have a written fire prevention plan.

Explanation:

Fires are unpredictable phenomena, since one never knows when they can be triggered. But if they can be prevented, and even more in work environments, through certain guidelines of precautions and basic care.

Fire prevention plans are necessary in all companies, and contain fire prevention measures and methods of fighting them when they have already been triggered. Thus, these plans not only provide basic prevention guidelines, but also behavior guidelines during fires: they teach how to use fire extinguishers, how to make orderly exits from the establishment, how to clear emergency exits, and how to activate alarms.

In short, these prevention plans are necessary as they make the safety of employees and the building, so they cannot and should not be overlooked.

kati45 [8]3 years ago

4 0

Answer:

True

Explanation:

if there is no prevention of fire, and when fire happens there will be many lives lost which is very sad so we need a prevention plan .

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Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types

zhenek [66]

Answer

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Step-by-step explanation:

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3 0

3 years ago

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve

Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature $T_1 = -16^0\ C$

Quality $x_1 = 0.2$

Outlet:

Temperature $T_2 = -16^0 C$

Quality $x_2 = 1$

The following data were obtained at saturation properties of R134a at the temperature of -16° C

$v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg$

$v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg$

$m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}$

$\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}$

3 0

3 years ago

A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =

oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0

3 years ago

Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and

yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

Solution:

- We will compute the velocity by integrating a by dt.

a = v*dv / dx = - (0.1 + sin(x/0.8))

- Separate variables:

v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

v = sqrt (0.104516)

v = +/- 0.323 m/s

- v = v_max when a = 0:

-0.1 = sin(x/0.8)

x = -0.8*0.1002

x = -0.080134 m

- Hence,

v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

v = sqrt (0.504)

v = +/- 1.004 m/s

4 0

3 years ago

soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new

timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

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We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0

3 years ago