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4 years ago

Give an example of a model code.

Engineering

1 answer:

Zanzabum4 years ago

6 0

Answer:

Model codes are the codes which used to change the location of a machine from one place to other place to perform specific task.g and M are the model codes which used in machines.

Ex:

G0 ,G1 , G2 ,G80 ,G81 ... For motion

G17 ,G18 , G19 For planer selection

Mo ,M1 , M2 ,M30 ,M60 For Stopping

M6 For tool change

M26 ,27 For clamping

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20 points and brainliest A, B, C, D

Annette [7]

B !! Is the correct answer

5 0

3 years ago

Read 2 more answers

A particular motor rotates at 3000 revolutions per minute (rpm). What is its speed in rad/sec, and how many seconds does it take

raketka [301]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

Now by putting the values in the above equation

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

$\dfrac{1}{50}=0.02\ s$

5 0

4 years ago

Read 2 more answers

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

An engineer who invested very well plans to retire now because she has $2,000,000 in her ORP account. How long will she be able

Ede4ka [16]

Answer:

20 years

Explanation:

You do 2,000,000 ÷ 100,000 but you can simplify that to 20 ÷ 1 = 20

During those 20 years, the profits you earn will be 80,000 since when you do 2,000,000 x 0.04 but you can simplify that to 20,000 x 4 getting 80,000 and that quite doesn't reach 100,000 dollars

8 0

3 years ago

In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en

Tasya [4]

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 = ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is 960 kg

6 0

4 years ago