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3 years ago

Interlocking stacked material is done by _____.

Engineering

2 answers:

VikaD [51]3 years ago

7 0

Answer:

B. Placing the objects at right angles

Explanation:

Workers are required to follow the safety guidelines when handling stacking materials. A material can fall or collapse causing injuries or death. According to OSHA guidelines, worker must do the following but not limited to:

when using hands, stacks should not exceed 16 feet high and 20 feet high if using a forklift
remove nails from used lumber before stacking
stacks should be stable and self-stable
material stored in racks should face away from main aisles

Dmitry [639]3 years ago

3 0

Answer:

Explanation:

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Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood

kakasveta [241]

Explanation:

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3 years ago

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

4 0

3 years ago

Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display

Citrus2011 [14]

Algorithm of the Nios II assembly program.

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

<h3>The Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally, the program will be written using a cpulator simulator in order to attain best result.

We are to

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0

2 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

3 years ago

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

<u>FOR SUBSTANCE A:</u>

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

<u>FOR SUBSTANCE B:</u>

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

5 0

3 years ago

Interlocking stacked material is done by​ _____.

Interlocking stacked material is done by _____.