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mr Goodwill [35]
4 years ago
11

If you are a government authority what extend will you modify the existing policy

Engineering
1 answer:
mr_godi [17]4 years ago
4 0

Answer:

kk

Explanation:

dkdndidodd ndidkjeeiwonejeeidmdnddkdidfmndd

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What should a technician do before entering a confined space? Question 1 options: A) Post another worker outside the confined sp
Lady bird [3.3K]

Answer:

C) Wear clothes that are resistant to spontaneous arcing.

Explanation:

A confined area or space is one that has strict restriction against movement of people due to the availability of hazardous substance or material within the confined space. If there is need for anyone to enter the space, it is very important that the person make use of personal protective equipment (PPE).

Personal protective equipment are materials that are capable of inhibiting injury or hazard when used appropriately. Examples are: helmet, respirators, gloves, goggles, ear muffs etc.

Before a technician should enter a confined space, he/ she should use a PPE by wearing clothes that are resistant to spontaneous arcing.

3 0
3 years ago
After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st
Mazyrski [523]
B evaluate the solution. Sorry if I'm wrong.
4 0
3 years ago
Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur
Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

<u>BP = 11995.2 Pa = 11.99 KPa</u>

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

7 0
3 years ago
The flowrate through a rectangular channel is 20 cfs. The upstream width of the channel is 10 ft, and the depth of the water in
Liula [17]

To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,

Q = 20cfs\\w= 10ft\\D= 1ft

Then the velocity would be:

V = \frac{Q}{wD}V = \frac{20}{10*1}V = 2ft/s

The number of Froude is given as,

Fr = \frac{V}{gD}^{1/2}

Where,

V = Velocity

g = Gravity

D = Diameter

Replacing we have that

Fr = \frac{2}{32.2}^{1/2}\\Fr = 0.352\\Fr

There will be no Jump, correct answer is B.

5 0
3 years ago
A desktop computer is to be cooled by a fan whose flow rate is 0.34 m3 /min. Determine the mass flow rate of air through the fan
Aleks [24]

Answer:

m = 0.238  <u>kg</u>

                  min

r = 0.063m

3 0
3 years ago
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