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2 years ago

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

Engineering

1 answer:

Liono4ka [1.6K]2 years ago

7 0

Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

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A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of

Kazeer [188]

This is the explanation

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3 years ago

Explain how feedback control is used to<br> adjust robotic movements.

LuckyWell [14K]

Answer:

Feedback control of arm movements using Neuro-Muscular Electrical Stimulation (NMES) combined with a lockable, passive exoskeleton for gravity compensation

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2 years ago

The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along

JulsSmile [24]

Answer:

D=41.48 ft

$a=54.43\ ft/s^2$

Explanation:

Given that

y=0.5 x²

Vx= 2 t

We know that

$V_x=\dfrac{dx}{dt}$

At t= 0 ,x=0

$x=\int V_x.dt$

At t= 3 s

$x=\int_{0}^{3} 2t.dt$

$x=[t^2\left\right ]_0^3$

x= 9 ft

When x= 9 ft then

y= 0.5 x 9² ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

$D=\sqrt{9^2+40.5^2} \ ft$

D=41.48 ft

$a_x=\dfrac{dV_x}{dt}$

Vx= 2 t

$a_x= 2\ ft/s^2$

At t= 3 s , x= 9 ft

y=0.5 x²

$a_y=\dfrac{d^2y}{dt^2}$

y=0.5 x²

$\dfrac{dy}{dt}=x\dfrac{dx}{dt}$

$\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}$

Given that

$\dfrac{dx}{dt}=2t$

$\dfrac{dx}{dt}=2\times 3$

$\dfrac{dx}{dt}=6\ ft/s$

$a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2$

$a_y=54\ ft/s^2$

$a=\sqrt{a_x^2+a_y^2}\ ft/s^2$

$a=\sqrt{2^2+54^2}\ ft/s^2$

$a=54.43\ ft/s^2$

7 0

3 years ago

Ethane (component A - C2H6) and hydrogen (component B) are fed to a differential reactor where they react on the catalyst to for

Fofino [41]

HELP ILL GIVE MOST BRAINLY AND 50 POINTS
HURRY PLEASE component c it is a compound so it will break

4 0

2 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

3 years ago