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Marrrta [24]
3 years ago
11

What is CAD/CAM (Computer-Aided Designing and Manufacturing) used for in school? Plz help

Engineering
2 answers:
joja [24]3 years ago
8 0
It used for designing homes and drafting products before production
jok3333 [9.3K]3 years ago
4 0

Answer:

CAD/CAM stands for computer-aided design & computer-aided manufacturing. CAD/CAM software is used to design and manufacture prototypes, finished products, and production runs.

Explanation:

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A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
Sliva [168]

Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

                  ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

   = \frac{144}{8}

   = 18 teeth/inch

The diametral is 18 teeth/inch

3 0
2 years ago
It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if KingHiero‘s cro
Rudiy27

Answer:

the crown is false densty= 12556kg/m^3[/tex]

Explanation:

Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.

W=mg

W=weight(N)=31.4N

M=Mass

g=gravity=9.81m/S^2

solving for M

m=W/g

m=\frac{31.4N}{9.81m/S^2}=3.2kg

The second step is find the volume of crown  remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

F=mg-\alpha  V g

Where

F=weight in water=28.9N

m=mass of crown=3.2kg

g=gravity=9.81m/S^2

α=density of water=1000kg/m^3

V= crown´s volume

solving for V

V=\frac{mg-F }{g \alpha } =\frac{(3.2)(9.81)-28.9}{9.81(1000)} =0.000254m^3

finally, we remember that the density is equal to the index between mass and volume

\alpha =\frac{m}{v} =\frac{3.2}{0.000254} =12556kg/m^3

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.

1.weigh the crown in the air and find the mass

2. put water in a cylindrical bucket and measure its height with a ruler.

3. Put the crown in the bucket and measure the new water level with a ruler.

4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.

5. find density by dividing mass by volume

7 0
3 years ago
Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo
bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0
2 years ago
Read 2 more answers
A team of engineers is designing a new rover to explore the surface of Mars. Which statement describes the clearest criterion fo
stich3 [128]

Answer:

hello im new trying to get points

Explanation:

3 0
2 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
2 years ago
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