Inspections may be_____ or limited to a specific area such as electrical or plumbing

Which examples best demonstrate likely tasks for Health, Safety, and Environmental Assurance workers? check all that apply

Over [174]

Sam, Elijah, Joy Those are 100% correct from my human knowledge

6 0

2 years ago

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7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21

Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

$g = (\frac{v}{s}) (\frac{C}{X})$

Let's first find the fourth stage critical lane group ratio $\frac{v}{s}$ , using the formula:

$C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}$

Solving for $(\frac{v}{s})$ , we have:

$(\frac{v}{s}) = 0.185$

Let's also calculate the volume capacity ratio X,

$X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}$

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

$g_4 = (\frac{v}{s}) (\frac{C}{X})$

Substituting figures in the equation, we now have:

$g_4 = (0.185) (\frac{60}{0.704})$

$g_4 = 15.78 seconds$

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0

2 years ago

The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg·°C. Express this value in kJ/kg·K, J/g·°C, kcal/ kg·°C, and Bt

Dmitry [639]

Answer:

a) C_v = 1.005 KJ/kgK

b) C_v = 1005.000 J/kgC

c) C_v = 0.240 kcal/kgC

d) C_v = 0.240 Btu/lbmF

Explanation:

Given:

- constant-pressure specific heat C_v = 1.005 KJ/kgC

Find C_v in units of:

a) kJ/kg·K

b) J/g·°C

c) kcal/ kg·°C

d) Btu/lbm·°F

Solution:

a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,

C_v = 1.005 KJ/kgK

b)

C_v = 1.005 KJ/kgC * ( 1000 J / KJ)

C_v = 1005.000 J/kgC

c)

C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)

C_v = 0.240 kcal/kgC

d)

C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)

C_v = 0.240 Btu/lbmF

6 0

3 years ago

A 50kg block of nickel at 90°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. Determine the fol

Valentin [98]

Answer:

a).Control volume

b). $C_{nickel}$ = 502.416 J/kg-K

c). $C_{water}$ = 4.187 kJ/kg-K

d). $T_{2}$ = 87.91°C

Explanation:

a). It is a control volume system because mass is varying in the system.

b). Specific heat of nickel is $C_{nickel}$ = 502.416 J/kg-K

c). Specific heat of water is $C_{water}$ = 4.187 kJ/kg-K

d).We know that

net energy transfer = change in internal energy

$m_{nickel}\times c_{nickel}(T_{2}-T_{nickel})=m_{water}\times c_{water}(T_{2}-T_{water})$

$m_{nickel}\times c_{nickel}(T_{2}-T_{nickel})=(volume_{water}\times density_{water})\times c_{water}(T_{2}-T_{water})$

$50\times 502.416\times (T_{2}-90)=(0.5\times 1000)\times 4.187\times (T_{2}-25)$

$25120.8\times (T_{2}-90)=2093.5\times (T_{2}-25)$

$T_{2}$ = 87.91°C

6 0

3 years ago

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Mobility refers to the ability to?

aev [14]

Answer:

Drive

Explanation:

Drive is a great example

4 0

3 years ago