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1 year ago

Why is it so dangerous to use a ground lift on a metal cased power tool

1 answer:

7 0

Answer:

Removal of the safety ground connection on equipment can expose users to an increased danger of electric shock and may contradict wiring regulations. ... If a fault develops in any line-operated equipment, cable shields and equipment enclosures may become energized, creating an electric shock hazard.

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a 250 pound person would use a type 1 ladder even if they were carrying a load with them true or faults

Marysya12 [62]

Answer:True

Explanation:

6 0

2 years ago

Read 2 more answers

Fill in the Blank: The _____ has a very sturdy handle with good leverage for tightening nuts and bolts. It can be used to apply

guapka [62]

Answer:

D Rachet

Explanation:

I tightens it

8 0

2 years ago

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

monitta

The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

3 0

2 years ago

For laminar flow over a flat plate, the local heat transfer coefficient hx is known to vary as x−1/2, where x is the distance fr

Reika [66]

Answer:

Explanation:

So for solving this problem we need the local heat transfer coefficient at distance x,

$h_x=cx^{-1/2}$

We integrate between 0 to x for obtain the value of the coefficient, so $\bar{h}_x =\frac{1}{x} \int\limit^x_0 h_x dx\\\bar{h}_x = \frac{c}{x} \int\limit^x_0 \frac{1}{\sqrt{x}}dx\\\bar{h}_x = \frac{c}{c} (2x^{1/2})\\\bar{h}_x = 2cx^{-1/2}$

Substituing

$\bar{h}_x=2h_x\\\frac{\bar{h}_x}{h_X}=2$

The ratio of the average convection heat transfer coefficient over the entire length is 2

6 0

2 years ago

The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely w

vodka [1.7K]

Answer:

The density of the unknown liquid is 1.025 kg/m³ (considering the density of the water as 1.000 kg/m³)

Explanation:

The hydrometer works by the Archimedes principle. The cylinder floats in the liquid because the hydrostatic thrust is equal to the weight force. This means:

$Tr-W=0N\\Tr=W\\\delta_{fl} \cdot Vol \cdot g =W_{hydr}$

If we measure 2 fluids, the weight of the hydrometer is the same, so:

$\delta_{fl1} \cdot Vol_1 \cdot g =W_{hydr}=\delta_{fl2} \cdot Vol_2 \cdot g\\\delta_{fl1} \cdot Vol_1 =\delta_{fl2} \cdot Vol_2\\\delta_{fl1} H_1 \pi R^2 =\delta_{fl2} H_2 \pi R^2\\\delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}$

If the original watermark height is 12.3cm (H₁) and the mark for water has risen 0.3 cm above the unknown liquid–air interface, the height of the unknown liquid mark is 12cm (H₂). Therefore:

$delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}\\1.000\frac{kg}{m^3} \frac{12.3cm}{12cm}=\delta_{fl2}=1.025\frac{kg}{m^3}$

8 0

2 years ago