These waveforms are applied to a gated D latch, which is initially RESET. Which of the areas identified on the Q waveform is inc

The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed

alex41 [359]

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed $\frac{\pi DL}{1000*f*V}$

Thus, we derive

Tc = $\frac{\pi (250)*625}{1000*0.3*V}$

Tc = $\frac{1636.25}{V}$

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × $\frac{1636.25}{V}$

According to the Taylor tool formula, cutting speed is expressed as

$VT^{0.25} = 700$

× V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

5 0

2 months ago

The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the

Kisachek [356]

Answer:

a) $F = 736.065\,lbf$ , b) $\mu_{k} = 0.15$

Explanation:

a) The uniform dresser can be modeled using specific equilibrium equations:

$\Sigma F_{x} = F - \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

Following some algebraic manipulations, the formulated equation is derived:

$F = \mu_{k}\cdot m \cdot g$

$F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )$

$F = 22.5\,lbf$

b) Similarly, the man can be represented by a set of equilibrium equations:

$\Sigma F_{x} = -F + \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

After some algebraic changes, the expression for the coefficient of static friction comes out as:

$\mu_{k} = \frac{F}{m\cdot g}$

$\mu_{k} = \frac{22.5\,lbf}{150\,lbf}$

$\mu_{k} = 0.15$

3 0

2 months ago

A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl

Kisachek [356]

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW

Explanation:

See explanation in the attached files

5 0

1 month ago

Compute the sum with carry-wraparound (sometimes called the one's complement sum) of the following two numbers. Give answer in 8

grin007 [323]

Response:

00100111

Explanation:

We have been given;

10010110

10010000

Add them following standard binary addition rules

10010110

10010000

-------------

(1)00100110

-------------

ignore the leading (1) because it is a carry.

Increase the result by 1 to achieve a 1's complement sum

00100110 + 1 = 00100111

Final Result: 00100111

3 0

2 months ago

Water flows steadily through a horizontal nozzle, discharging to the atmosphere. At the nozzle inlet the diameter is D1; at the

mote1985 [299]

The inlet gauge pressure must be 61.627 Psi.

4 0

2 months ago