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erastova [34]
3 years ago
6

Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect

ion handout, the class was called Date, but we have renamed it here because Practice-It uses a class named Date for other purposes.) The class includes the following members:

Engineering
1 answer:
laiz [17]3 years ago
7 0

Answer:

note:

find the attachment

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Side milling cutter is an example of ______ milling cutter.
dusya [7]

Answer:

special type

Explanation:

As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.

6 0
3 years ago
Which of the following has special properties that allow forces and pressure to be distributed evenly?
Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

  • Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
  • Therefore, when pressure is applied to them, it permeates evenly on all parts.
  • Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.
7 0
3 years ago
Indicates the design of the building and<br> adjoining areas
Lelu [443]

Answer:

Architectural Plan

Explanation:

4 0
3 years ago
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Which of these cars traveled faster during time interval <br> please show solution
Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0
3 years ago
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