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2 years ago

Elements of parallel computing

Engineering

1 answer:

Leya [2.2K]2 years ago

6 0

$\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}}$

<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

Computer systems organization.
Computing methodologies.
General and reference.
Networks.
Software and its engineering.
Theory of computation.

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Tech A says that both OSHA and the EPA can inspect facilities for violations. Tech B says that a shop safety rule does not have

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Answer:

A is right, depending on what you mean the health organization OSHA is aloud to search faculty if they can or may be affecting peoples health. With a proper search it can be made clear if its safe or not and the OSHA is aloud to do it without your consent to it.

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3 years ago

8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of

san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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3 years ago

The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, d

Anna007 [38]

Answer:

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s\\$

In hp:

$p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp$

Part B:

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s\\$

In hp:

$p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp$

Explanation:

Weight of elevator=1000-lb

Force=500 lb

s=15 ft

Force on pulley=F=3*500=1500 lb

g= $32.2 ft/s^2$

According to Newton's Second law:

$\sum F_y=ma_y$

According to attachment:

$F-W=ma_y$

$1500-1000=\frac{1000}{32.2}a_y$

$a_y=16.1 ft/s^2$

According to third equation of motion:

$v^2=v_o^2+2a_y(S-So)\\$

Where:

Vo is initial velocity

V is final velocity

S is final distance

So is starting distance

$v^2=(0)^2+2*16.1*(15)\\v^2=483\\v=21.977 ft/s$

Output power:

$P_{out}=F.v\\P_{out}=1500*21.977\\P_{out}=32965.5 lb.ft/s$

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s\\$

In hp:

$p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp$

Part B:

When S=35 ft

$v^2=(0)^2+2*16.1*(35)\\v^2=1127\\v=33.5708 ft/s$

Output power:

$P_{out}=F.v\\ P_{out}=1500*33.5708 \\ P_{out}=50356.2 lb.ft/s$

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s\\$

In hp:

$p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp$

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3 years ago

What advantage might there be to having the encoder located on the motor side of the gearhead instead of at the output shaft of

pickupchik [31]

The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.

Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.

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4 years ago