Please help ill mark as brainlest

In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in

stellarik [79]

Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

T = 5 minutes

F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

$T = \frac{pi* D * L}{V*F}$

We are to find the speed, V. Let's make it the subject.

$V = \frac{pi* D * L}{F*T}$

Substituting values we have:

$V = \frac{pi* 0.4 * 0.15}{0.0003*5}$

V = 125.68 m/min ≈ 125.7 m/min

Therefore, V = 125.7m/min

7 0

3 years ago

Assuming the point estimate in Problem 6.36 is the true population parameter, what is the probability that a particular assay, w

sp2606 [1]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $0.958$

Explanation:

The explanation is shown on the second and third uploaded image

3 0

3 years ago

Introduce JTA and JT

Vinil7 [7]

JT2 oaoaosnforeneomdocdmsnsksmsmsodnfnfj

7 0

2 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

where D is constant

then

$x^2$ / t = constant

$x^2_1$ / t₁ = $x^2_2$ / t₂

$x^2_1$ t₂ = t₁ $x^2_2$

t₂ = t₁ $x^2_2$ / $x^2_1$

t₂ = ( $x^2_2$ / $x^2_1$ )t₁

t₂ = $($ $x_2$ / $x_1$ $)^2$ × t₁

so we substitute

t₂ = $($ 0.0049 / 0.0018 $)^2$ × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0

2 years ago

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp

cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

6 0

3 years ago