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Leokris [45]
2 years ago
11

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

Engineering
1 answer:
pantera1 [17]2 years ago
3 0

Answer:

Eye.

Explanation:

Dust, dirt, and metal chips are most unpleasant to get in your eyes. Just experience it and you'll know what I mean.

;)

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The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr
Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below;  if we take an integral look at the component of force in the direction of the displacement; we have

F_x= Fcos 40°

F_x= 100 (cos 40°)

F_x= 76.60 N

Workdone by the friction force can now be determined as:

W = F_x × displacement

W = 76.60 × 2

W = 153.2 J

∴  the work done by the friction force = 153.2 J

7 0
2 years ago
A fuel oil is burned with air in a furnace. The combustion produces 813 kW of thermal energy, of which 65% is transferred as hea
DENIUS [597]

Answer:

im not sure

Explanation:

8 0
2 years ago
Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor ack
Elza [17]

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

3 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
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