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2 years ago

Excepted packaging refers to fiberboard boxes, sturdy wooden boxes, or steel crates that are designed to transport:

1 answer:

5 0

Excepted packaging refers to fiberboard boxes, sturdy wooden boxes, or steel crates that are designed to transport materials with extremely low levels of radioactivity.

<h3>What is excepted packaging?</h3>

Excepted packaging can be defined as a type of box that is designed and developed for the transportation of materials with extremely low levels of radioactivity.

This ultimately implies that, a type of box which is strictly used for the transportation of materials with extremely low levels of radioactivity in order to prevent hazard.

Read more on packaging here: brainly.com/question/14985175

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Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the

Alex_Xolod [135]

Answer:

$P_{2} = 527.803\,kPa$

Explanation:

The politropic relationship for a isentropic process is:

$\frac{P_{2}}{P_{1}} = \left(\frac{V_{1}}{V_{2}} \right)^{\gamma}$

Where $\gamma$ is the ratio of specific heats

The final pressure is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$

$P_{2} = (200\,kPa)\cdot (2)^{1.4}$

$P_{2} = 527.803\,kPa$

7 0

3 years ago

What is the resolution limit for a projection type of photolithographic system if the incident wavelength is 365 nm (the i-line

Vadim26 [7]

Answer:

Depth of field = 347.619 nm

Explanation:

wavelenght = 365nm

N.A =0.63

k1= 0.6

so we have that the resolution limit is:

R=k1*A

R=0.6*365

R=219 nm

and the Depth of field needed for the best resolution is:

DoF = Resolution / N.A.

DoF= R/N.A

DoF= 219/0.63

DoF= 347.619 nm

7 0

3 years ago

A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan

hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

$R = \dfrac{u^2}{15(e+f_s)}$

where;

R= radius

$f_s$ = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

$f_s$ = 0.20

So;

$R = \dfrac{30^2}{15(0.08+0.20)}$

$R = \dfrac{900}{15(0.28)}$

$R = \dfrac{900}{4.2}$

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0

3 years ago

Vehicles begin arriving at an amusement park 1 hour before the park opens, at a rale of four vehicles per minute. The gale to th

Alina [70]

Answer:

a) ≈ 30 mins

b) 8 vpm

Explanation:

<u>a) Determine how long after the first vehicle arrival will the queue dissipate</u>

The time after the arrival of the first vehicle for the queue to dissipate

= 29.9 mins ≈ 30 mins

<u>b) Determine the average service rate at the parking lot gate </u>

U = A / t

where : A = 240 vehicles , t = 30

U = 240 / 30 = 8 Vpm

attached below is a detailed solution of the given problems above

3 0

3 years ago

During the reaction, 3.50 μmol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is comp

lyudmila [28]

Answer:

The pH of the solution will be equal to 5.46

Explanation:

The dissociation reaction of HCl is equal to:

HCl → H+ + Cl-

To solve the exercise we must first convert the µmoles to moles using the following conversion factor:

3.5µmoles x $\frac{1 mol}{1x10^{6} umol}$ = 3.5x $10^{-6}$ moles

Assuming a liter of solution, we can calculate the molar concentration by:

M = $\frac{Number of moles}{Liter of solution}$

Replacing:

M = $\frac{3.5x10^{-6}moles }{1 L}$ = 3.5x $10^{-6}$ moles/L

As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5x $10^{-6}$ moles/L

The pH will be equal to:

pH = -log[H+]

Replacing:

pH = -log[3.5x $10^{-6}$ ] = 5.46

8 0

4 years ago