Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood

PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs

iogann1982 [368]

Answer:

?pooooooooooooooooooooooop

Explanation:

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6 0

2 months ago

A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom where water discharges to the atmosphere. Th

grin007 [323]

Answer:

A fluid is a material that maintains continuous and permanent deformity when a shearing stress is applied.

• The pressure at a given point in a fluid remains unaffected by the direction of the surface that intersects this point; this pressure is isotropic.

• The force generated by pressure p acting on one side of a minimal surface area dA defined by a unit normal vector n can be expressed as −pndA.

• The speed at which pressure is conveyed through a fluid matches the speed of sound.

• The units employed vary based on the selected system, incorporating measures such as feet, seconds, newtons, and pascals. On the other hand, a dimension refers to a more abstract idea, encompassing terms like mass, length, and time.

• The specific gravity (SG) of a solid or liquid is the proportion of its density compared to that of water at the identical temperature.

• A Newtonian fluid is characterized by the viscous stress being directly proportional to the strain rate (velocity gradient). The viscosity, µ, is a property of the fluid that varies with temperature.

• At the boundary between solid and fluid, the velocities of both the fluid and solid coincide; this situation is referred to as the “no-slip condition.” Consequently, with high Reynolds numbers (>> 1), boundary layers develop near the solid surface. In those layers, significant velocity gradients occur, resulting in crucial viscous influences.

• At the junction of two distinct fluids, surface tension may play a significant role. Surface tension leads to the emergence of phenomena like meniscus, drops, bubbles, and capillary rise seen in narrow tubes since it can counterbalance pressure variations across the interface.

5 0

1 month ago

Who want to play "1v1 lol unblocked games 76"

Daniel [329]

Answer:I want to know what game to play?

Explanation:

3 0

3 months ago

Read 2 more answers

The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re

mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, $g = 9.81 m/s^2$

mass flow rate, $\dot{m} = 60.6 kg/s$

flow density, $\rho = 1000 kg/m^3$

efficiency of the pump, $\eta = 0.74$

output gauge pressure, $p_o = 344.75 kPa$

input gauge pressure, $p_i = 68.95 kPa$

cross-sectional area of output pipe, $A_o = 0.069 m^2$

cross-sectional area of input pipe, $A_i = 0.093 m^2$

height of discharge, $z_o = 1.22 m - 0.61 m = 0.61 m$ (evaluated at pump’s maximum height of 1.22 m)

input height, $z_i = 0 m$

hydraulic power of the pump, $P =? kW$

Initially, the volumetric flow (Q) needs to be determined

$Q = \frac{\dot{m}}{\rho}$

$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

$Q = 0.0606 m^3/s$

Next, compute the velocity (v) for both input and output

$v_o = \frac{Q}{A_o}$

$v_o = \frac{0.0606 m^3/s}{0.069 m^2}$

$v_o = 0.88 m/s$

$v_i = \frac{Q}{A_i}$

$v_i = \frac{0.0606 m^3/s}{0.093 m^2}$

$v_i = 0.65 m/s$

Subsequently, the total head (H) can be calculated

$H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}$

$H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}$

$H = 28.74m$

Finally, the computation of pump power is done as follows

$P = \frac{Q \, \rho \, g \, H}{\eta}$

$P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}$

$P = 23.09 kW$

6 0

2 months ago

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