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3 years ago

Select all that apply: Contaminated sharps should not be ----

Engineering

1 answer:

PIT_PIT [208]3 years ago

4 0

Answer:

Contaminated sharps should not be bent, recapped or removed.

Explanation:

Contaminated sharps are defined as "any contaminated object that can penetrate the skin including, but not limited to, needles, scalpels, broken glass, broken capillary tubes and exposed ends of dental wires".

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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

4 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 106 GPa and an original cross-sectional diameter of 3.9

kiruha [24]

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL = $\dfrac{FL}{AE}$ ....................1

put here value and we get L

L = $\dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}$

solve it we get

L = 0.312752 m

L= 312.75 mm

8 0

4 years ago

Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/

kap26 [50]

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .

Sensible heat for water Q

$Q=mC_p\Delta T$