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21 day ago

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A technician has been dispatched to assist a sales person who cannot get his laptop to display through a projector. The technici

an verified the video is displaying properly on the laptop's built-in screen. Which of the following is the next step the technician should take?

1 answer:

mote1985 [204]21 day ago

7 0

Answer:Ensure the correct cable is connected between the laptop and the projector. Check for HDMI inputs or 15-pin video output interfaces.

Also, make sure the laptop is set to project to the correct display output.

Explanation:

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1. A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions

choli [191]

Answer:

The highest vehicle count = 308

Explanation:

Refer to the attached document for the calculations.

6 0

1 day ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

mote1985 [204]

Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$ r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Next by substituting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

Concluding, the shearing angle is 18.03°.

4 0

1 month ago

6. You are evaluating flow through an airway. The current flow rate is 10 liters per minute with a fixed driving pressure (P1) o

iogann1982 [279]

Answer:

B) P1 would have to increase to sustain the flow rate (correct)

C) Resistance would rise (correct)

Explanation:

Flow rate is measured at 10 liters per minute

Driving pressure (P1) stands at 20 cm H2O

Fixed downstream pressure (P2) is 5 cm H2O

The accurate statements when the lumen is pinched in the center of the tube are: P1 will increase to maintain the flow rate, and resistance will rise. This occurs because pinching the lumen decreases its diameter, leading to higher resistance, which is linearly related to pressure, thus P1 will also increase.

The incorrect statement is: the flow would decrease.

6 0

16 days ago

The top 3 most popular male names of 2017 are Oliver, Declan, and Henry according to babynames. Write a program that modifies th

Viktor [230]

Answer:

The Python code is provided below with suitable comments

Explanation:

# convert list to set

male_names = set(['Oliver','Declan','Henry'])

# get name for removal and addition from user

remove_name = input("Enter remove name: ")

add_name = input("Enter add name: ")

# remove specified name from set

male_names.remove(remove_name)

# add new name to set

male_names.add(add_name)

# sort the resulting set

a = sorted(male_names)

# output the sorted set

print(a)

7 0

1 month ago

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 [204]

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

25 days ago