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Andrej [43]
3 years ago
10

Power supplies used in desktop computers many years ago, commonly referred to as AT, had a power switch that switched on the 120

VAC going into the power supply. Today’s desktop computers, commonly referred to as ATX, use a power switch that sends a low-voltage logic signal to the _____.
Engineering
1 answer:
Karolina [17]3 years ago
4 0

Answer:

Power supply relay coil

Explanation:

The power supply known as ATX used in today's PCs is very complicated kind of power supply. The main objective of ATX form of the power supply is conversion of the input supply from the line into different voltages to be supplied in order to run the desktop computer's mother board.

While the AT supply that was used many years back that supplied about 120 VAC, The ATX power supply in today's world is capable of supplying power supply of about 300 W or more.

Power supply relay coil is a solitary estimation of source voltage planned by configuration to be connected to the input or coil.

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A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a componen
jekas [21]

Answer:

26.7 min

Explanation:

First, we will find the <u>time required to drill each hole</u>:

  • N = 300 x 12/0.75 \pi = 1527.7 rev/min
  • fr = 1527.7 (0.015) = 22.916 in/min

Formula for <u>distance per hole</u>: 0.5 + A + 1.75

  • A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
  • Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min

Now, we will calculate the <u>time required to draw back the drill form hole</u>:

              = 0.112 / 2 = 0.056 min

Time to move between holes = 1.5 / 15 = 0.1 min

For 100 holes, the number of moves between holes = 99

Total time required to drill 100 holes (t):

                       t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min

7 0
3 years ago
Read 2 more answers
Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo
almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) v_{y}

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) v_{y}

⇒ 30 + 4(6) = 4.05 v_{y}

⇒30 +24 = 4.05 v_{y}

⇒54 = 4.05 v_{y}

⇒v_{y} = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + v_{y}² = √(10.26)² + (13.33)²

                                                           = √105.26 + 177.68

                                                           = √282.95 = 16.82

The angular direction, θ =  tan^{-1}(\frac{v_{y} }{v_{x} }) =  tan^{-1}(\frac{13.33}{10.26}) = tan^{-1}(1.299) = 52.41°

8 0
3 years ago
"geophysical exploration definition"​
Strike441 [17]

Answer:

Exploration geophysics is an applied branch of geophysics and economic geology, which uses physical methods, such as seismic, gravitational, magnetic, electrical and electromagnetic at the surface of the Earth to measure the physical properties of the subsurface, along with the anomalies in those properties

7 0
3 years ago
Read 2 more answers
Steam at 4 MPa, 400°C enters a steady-flow, adiabatic turbine through a 20 cm-diameter-pipe with a velocity of 20 m/s. It leaves
Furkat [3]

Answer:

28,8 m/s

Explanation:

In a steady flow system we can say that m1=m2 which means that the mass flow in the entrance in the same in the outlet.  m is flow (kg/s)

we know that m=\frac{1}{v} A*V where V (m/s) is velocity, A (m^2) ia area and v is specific volume (m^3/kg)

Since m1=m2 we can say

\frac{1}{v_{1} } A_{1} *V_{1}=\frac{1}{v_{2}} A_{2} *V_{2}

clearing the equation

V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}

we can specific volume (m^3/kg) from thermodynamic tables

for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg

In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated  (vg) gas from the thermodynamic tables. we use the next equation to get  (v) for the condition of interest, in this case 80% quality.

v= vf +x*(vg - vf)

where:

x: quality

vf = liquid-saturated-specific-volume

vg =steam-saturated-specific-volume.

for this problem

x = 0,8

vf = 0,00102991

vg = 3,24015

so

we get = 2,593 m^3/kg

The area is the one for a circle

\pi *r^{2}

r1 = 0,1 m^2 for area 1

r2=0,5 m^2 for area 2

A1 = 0,0314 m^2

A2 = 0,7853 m^2

we know that  V1 is 20 m/s

replacing these values in the equation

V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}

we get V2 = 28,2 m/s.

3 0
3 years ago
Can somebody please help me
astraxan [27]

Answer:

C. 420 x 594 millimeters

6 0
3 years ago
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