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wolverine [178]
3 years ago
13

What is the command this line of code is telling the robot?

Engineering
2 answers:
podryga [215]3 years ago
4 0

Answer:

drive the car

Explanation:

Yuki888 [10]3 years ago
4 0
Drive Forward or move forward.
You might be interested in
Ratio equivalent to 12 red beans to 5 total beans
Kruka [31]

Hello!

The answer to this question as a percentage is 240%

8 0
3 years ago
A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of ins
Stella [2.4K]

Answer:

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

Explanation:

Generally, thermal resistance for conduction heat transfer in a sphere.

R_{cond} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi K}}  

Where R_{cond} is the thermal resistance for conduction, K is the thermal conductivity of the material, r_{i} is the inner radius of the sphere, and r_{o} is the outer radius of the sphere.

The surface area of sphere, A_{s} is given by

A_{s}=4\pi {r^2}

For aluminum sphere, the thermal resistance for conductive heat transfer is given by

Calculate the thermal resistance for conductive heat transfer through the aluminum sphere.

R_{cond,s{\rm{ - 1}}} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}}

Where K_{Al} is aluminum’s thermal conductivity at T_{s}

Thermal resistance for conductive heat transfer through the insulation.

R_{cond,1{\rm{ - 2}}} = \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}}

Thermal resistance for convection is given by

R_{conv} = \frac{1}{{hA}}

Where h is convective heat transfer coefficient, R_{conv} is thermal resistance for convection and A is the cross-sectional area normal to the direction of flow of heat energy

Thermal resistance for convective heat transfer in-between the outer surface of the insulation and the ambient air.

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h{A_s}}}

Where h represents convective heat transfer coefficient at the outer surface of the insulation. Since A_{s} is already defined, substituting it into the above formula yields

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h\left( {4\pi {r^2}} \right)}}

To obtain radial distance of the outer surface of the insulation from the center of the sphere.

r = r_{o} + t where t is thickness of insulation

r=0.21+0.15=0.36m

Total thermal resistance

R_{eq} = {R_{cond,s{\rm{ - 1}}}} + {R_{cond,1{\rm{ - 2}}}} +{R_{conv,2{\rm{ - }}\infty }}

Where R_{eq} is total thermal resistance

R_{eq} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}

Consider the thermal conductivity of aluminum at temperature T_{s} as 234W/m.K

Rate of heat transfer for the given process

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{{R_{eq}}}}

Where \dot Q_{s - \infty }} is the steady state heat transfer rate in-between the inner surface of the sphere and the ambient air.

Substituting \left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right) for R_{eq} we obtain

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)}}

\begin{array}{l}\\80{\rm{ W}} = \frac{{250{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}}}{{\left( {\frac{{\left( {\frac{1}{{0.18{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.21{\rm{ m}}}}} \right)}}{{4\pi \left( {234{\rm{ W/m}} \cdot {\rm{K}}} \right)}} + \frac{1}{{30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {4\pi {{\left( {0.36{\rm{ m}}} \right)}^2}} \right)}}\frac{{\left( {\frac{1}{{0.21{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.36{\rm{ m}}}}} \right)}}{{4\pi {K_{ins}}}} + } \right)}}\\\\80{\rm{ W}}\left( {{\rm{0}}{\rm{0.020737 K/W}} + \frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}}} \right) = 230{\rm{ K}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = \frac{{230{\rm{ K}}}}{{80{\rm{ W}}}} - {\rm{0}}{\rm{0.020737 K/W}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = {\rm{2}}{\rm{.854263 K/W}}\\\end{array}

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

7 0
4 years ago
1) What output force (Fout) is produced if the lever arm length (rout) is 100 mm?
Ierofanga [76]
The length of the arm is the main part of natur
6 0
3 years ago
Read 2 more answers
Question Completion Status:
Mashutka [201]

Answer: c. Centre of pressure​

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

6 0
3 years ago
You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen
VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

6 0
3 years ago
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