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3 years ago

On a CNC lathe, which of the following axes specifies the distance of the cutting tool from the workpiece centerline?

Engineering

1 answer:

kvasek [131]3 years ago

8 0

Answer:

The axis of motion that is parallel to the spindle axis is always the Z-axis.

Explanation:

Z-Axis Which axis is which depends on the orientation of the spindle.

If the spindle is vertical (Figure 2.1), the Z-axis is vertical. Either the quill or the knee of a vertical spindle mill will move when a Z-axis command is executed.

This is the best answer I could give you, maybe you could show us a pic of the question so we see all the possible choices?

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. What are the benefits of synthetic motor oil compared to conventional oil?

madam [21]

Explanation:

the answet is E. all of the above

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3 years ago

What’s Statistics<br> What are the 2 Source of error in data collection

FromTheMoon [43]

Answer:

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3 years ago

Read 2 more answers

Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F.

zvonat [6]

Answer:

In Rankine 524.07°R

In kelvin 291 K

In Fahrenheit 64.4°F

Explanation:

We have given temperature 18°C

We have to convert this into Rankine R

From Celsius to Rankine we know that $T(R)=(T_{C}+273.15)\frac{9}{5}$

We have to convert 18°C

So $T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R$

Conversion from Celsius to kelvin

$T(K)=(T_{C}+273)$

We have to convert 18°C

$T(K)=(18+273)=291K$

Conversion of Celsius to Fahrenheit

$T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F$

7 0

3 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.

Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature $T_1=8^{\circ}C$

$T_2=28^{\circ}C$

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

$Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ$

We know that $COP=\frac{Q_1}{W}$

$1.5=\frac{Q_1}{450}$

$Q_1=675$ W=0.675 KW

so time required to cool the watermelon is

$t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec$

4 0

3 years ago