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musickatia [10]
2 years ago
6

The amount of straight pipe needed to bend a given radius. Also, the actual length of the conduit that will be bent.

Physics
1 answer:
slava [35]2 years ago
8 0

Development length: The actual length of the bent conduit. Gain: Since a conduit bends radially rather than at an angle, the total length will not match the length required for all bends. Gain is the amount of space that is saved by a 90^{0} curve.

<h3>Bent Conduit</h3>

Conduit benders from Klein Tools are built to function and last longer than even the highest professional standards. To ensure a favourable experience and significantly enhance the final result of your project, it is advised that you become familiar with bending concepts, procedures, and the bender's capabilities. The benders are labelled with various alignment symbols to enable the operator make the bends required to complete any job. This aids bending while executing a ground or air bend. Arrow, teardrop, star point, and angle markings are the symbols on the Klein Tools benders. On certain bender head sides, you can see these markings.

Learn more about bent conduit here:

brainly.com/question/5023977

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A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
3 years ago
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