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olasank [31]
2 years ago
13

If I have a scaffold that is 83' tall, and the base of my scaffold is 5' wide least base dimension, then I am allowed to space m

y intermediate bracing supports every ____ feet maximum vertically with the lowest brace being no greater than _____ feet off the ground.
Engineering
1 answer:
Minchanka [31]2 years ago
8 0

In the scaffold, the intermediate bracing supports every 26 feet while the the lowest brace being no greater than 20 feet.

<h3>What is scaffold?</h3>

Scaffolding simply means a temporary structure that's used to support a work crew.

In this case, the height is 83 feet and the base width is 5 feet. The height to the width ratio will be:

= 83/5 = 16.6 > 4.

Therefore, the scaffold will be restrained. Lowest brace will be placed at 20 and intermediate support braces at 26 feet.

Learn more about scaffolding on:

brainly.com/question/20726451

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I just wanted to say thanks you to Brainly. This website is a huge help!
Fantom [35]

THANKS BRAINLY <3

Explain : I've got a lot of help from this website :D

3 0
3 years ago
All machines have three fundamental hazards: moving parts, point of operation, and?
OlgaM077 [116]

Answer:

All machines have three fundamental hazards: moving parts, point of operation, and the power transmission.

Explanation:

The unit that supplies power to the machine is a critical hazard due to high energy sources being potential fatal if proper protocols are not followed. This is why lockout tagout (LOTO) measures are put in place in order to protect people while they work on equipment.

3 0
2 years ago
What are the relative volume requirements for a CM reactor compared to a PF reactor if first-order kinetics apply and treatment
Tanzania [10]

Answer:

look it up

Explanation:

6 0
3 years ago
A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
Why are diode logic gates not suitable for cascading operation?
sasho [114]

Diode logic gates are not suitable for cascading operation because the pull-down or pull-up resistance within makes them a high output resistance device.

<h3>What is a cascading operation?</h3>

A cascading operation can be defined as a process which involves the use of a circuit breaker's current-limiting capacity in order to enable an installation of lower-rated and lower-cost circuit breakers.

According to the National Electrical Code (NEC), diode logic gates are not suitable for cascading operation because the pull-down or pull-up resistance within makes them a high output resistance device.

Read more on National Electrical Code here: brainly.com/question/10619436

#SPJ1

5 0
2 years ago
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