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Lostsunrise [7]
3 years ago
6

How to walk a dog dududududududududesssss

Engineering
2 answers:
bagirrra123 [75]3 years ago
6 0
Pick up a leash put the dog on the leash open the door/cage take the dog out and walk it
rusak2 [61]3 years ago
4 0
Putting them on a leash and picking a route and walking it
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Which of the following best describes the relationship between the World Wide Web and the Internet? А The World Wide Web is a pr
Gwar [14]

Answer:

C

Explanation:

5 0
2 years ago
Which of following are coding languages used in controlling a robot? *
Bess [88]

Answer:

C/C++

Explanation:

C/C++

7 0
2 years ago
A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find
Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

\sigma=\frac{\textup{PD}}{\textup{2t}}

where, t is the thickness

on substituting the respective values, we get

100=\frac{\textup{2\times800}}{\textup{2t}}

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0
3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
3 years ago
............ ..........<br>​
sukhopar [10]

Answer:

...................

Explanation:

6 0
2 years ago
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