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3 years ago

Name 3 ways in which robots have improved since the Ebola outbreak.

Engineering

1 answer:

Ostrovityanka [42]3 years ago

4 0

Answer:

1. They have less malfunctions

2. They can be operated from afar

3. Some are self-operated.

Explanation:

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The acceleration (in m/s^2) of a linear slider (undergoing rectilinear motion) within a If the machine can be expressed in terms

Inga [223]

Answer:

47.91 sec

Explanation:

it is given that $\alpha =\frac{1}{4v^{2}}$

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 $\frac{m}{sec^{2}}$

we know that $\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}$

$4v^{2}dv=dt$

integrating both side

$\frac{4v^{3}}{3}=t+c$ ---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so $\frac{4v^{3}}{3}=t$

when v= 3.3 $\frac{m}{sec^{2}}$

t= $\frac{4}{3}\times 3.3^{3}$

=47.91 sec

6 0

4 years ago

Millions of years ago, the Sierra Nevada region began to be uplifted along a crack in Earth's crust. The region on the other sid

Evgen [1.6K]

Answer: The correct answer is : Fault block mountain with rough edges and steep cliffs

Explanation: Snowy saws are an example of a mountain chain blocked by faults. The snowy mountains were formed because the tectonic movement forced some segments of the earth's crust up into irregular pieces and others down.

8 0

4 years ago

Since the passing of the Utah GDL laws in 1999:

wlad13 [49]

The answer is b, I hope this helps you

7 0

3 years ago

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

4 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago