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3 years ago

How far away from the compressor unit should an air transformer be installed?

Engineering

1 answer:

Setler [38]3 years ago

4 0

Answer:

During installation, the air transformer should be located at least 25 feet minimum from the compressor unit

Explanation:

An air transformer is used to remove the dirt, water and oil in the compressed air and to filter as well as to provide air regulation and are required in spray finishing tasks that calls for a regulated clean and dry supply of air

The air transformer measures the pressure of the regulated air and can also come with air outlets for air operated tools, blow guns, spray guns, as well as for other purposes

In order to operate efficiently, the air transformer are required to be installed at a minimum distance of 25 feet (7.6 m) from the compressor unit.

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Answer:

Not having enough money to pay to get it fixed

Explanation:

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Write definitions for the following two functions:______

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Answer:

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def SumN(n):

total = 0

for i in range(1,n+1):

total = total + i

print("Total: ",total)

def SumNCubes(n):

total = 0

for i in range(1,n+1):

total = total + i**3

print("Cube: ",total)

n = int(input("User Input: "))

if n > 0:

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Explanation:

The SumN function is defined here

def SumN(n):

This line initializes the total to 0

total = 0

The following iteration compute the required sum

for i in range(1,n+1):

total = total + i

This line outputs the calculated sum

print("Total: ",total)

The SumNCubes function is defined here

def SumNCubes(n):

This line initializes the total to 0

total = 0

The following iteration compute the required sum of cubes

for i in range(1,n+1):

total = total + i**3

This line outputs the calculated sum of cubes

print("Cube: ",total)

The main starts here; The first line prompts user for input

n = int(input("User Input: "))

The next line checks if input is greater than 0; If yes, the two defined functions are called

if n > 0:

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3 years ago

A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 90% and the hit rate of the level 2 cache

Hatshy [7]

Answer:

B) 2.22

Explanation:

In a first step, the system will look in cache number 1. If it is not found in cache number 1, then the system will look in cache number 2. Finally (if not in cache 2 also) the system will look in main memory.

The average access time will take into consideration success in cache 1, failure in cache number 1 but success in cache number 2, failure in cache number 1 and 2 but success in main memory.

Mathematically we can write it into the following equation :

$AAT=(H1.T1)+(1-H1).H2.T2+(1-H1).(1-H2).Hm.Tm$

Where AAT is the average access time

H1,H2 and Hm are the hit rate of cache 1,cache 2 and main memory respectively.

T1,T2 and Tm are the access time of cache 1, cache 2 and main memory respectively

Hm = 1

$AAT=(0.9).(1)+(1-0.9).(0.8).(4)+(1-0.9).(1-0.8).(1).(50)=0.9+0.32+1=2.22$

$AAT=2.22$

Therefore, option b) is the correct.

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3 years ago

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Can anyone answer this question

Solnce55 [7]

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

$t = \sqrt{\frac{2*X(t)}{a}}$

$t = \sqrt{\frac{2*18}{2.81}} = 3.6 s$

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

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4 years ago