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3 years ago

11

use the emu8086 to write a program for pushing AX seven times at the stack, check the stack pointer. and [hint : initialize with

8086H] .

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

I got the point

Explanation:

just chill out baby

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You are the project manager assigned to construct a new 10-story office building. You are trying to estimate the costs for this

Semmy [17]

Answer:

Bottom-up Estimation

Explanation:

Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.

This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.

8 0

3 years ago

The flowrate through a rectangular channel is 20 cfs. The upstream width of the channel is 10 ft, and the depth of the water in

Liula [17]

To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,

$Q = 20cfs\\w= 10ft\\D= 1ft$

Then the velocity would be:

$V = \frac{Q}{wD}V = \frac{20}{10*1}V = 2ft/s$

The number of Froude is given as,

$Fr = \frac{V}{gD}^{1/2}$

Where,

V = Velocity

g = Gravity

D = Diameter

Replacing we have that

$Fr = \frac{2}{32.2}^{1/2}\\Fr = 0.352\\Fr$

There will be no Jump, correct answer is B.

5 0

3 years ago

Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5

pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$

$Pb = 70000\ Pa-27303\ Pa - 14715\ Pa$

$Pb = 27,996\ Pa = 28\ kPa$

6 0

3 years ago

to determine cam ring speed you must use a ___________________,____________________or a________________

saul85 [17]

Answer:

=> The total numbers of cylinders on the engine.

=> Total number of lobes in the cam ring.

=> The direction at which the cam ring rotates.

Explanation:

A cam is a kind of ring and a device that is being used in engines. One or the main purpose of using cams in engines us because it helps in the change or transformation of the rotational movement of the engine to a translational one. Instead of using a cam ring, a cam roller can be used in place.

There are three things that can be used to to determine the speed of cam ring speed and they are given below as;

=> The total numbers of cylinders on the engine.

=> The Total number of lobes in the cam ring.

=> The direction at which the cam ring rotates.

5 0

4 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

3 years ago