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Ganezh [65]
1 year ago
10

If a bender can be used to bend 3/4-inch rmc, then it can also be used to bend__________emt.

Physics
1 answer:
USPshnik [31]1 year ago
3 0

A bender may also be used to bend 1-inch emt if it can bend 3/4-inch rmc.

<h3>A hand bender is what?</h3>

hand bender. a device with a diameter of no more than 1-1/4 in. that is used to bend EMT. To prevent crimping or flattening the conduit during bending, hand benders offer a radius that supports the conduit.

<h3>When would you utilize a hand bending device?</h3>

The user can produce very close bends with hockeys because they are 60% smaller than benders. Generally speaking, you should use a hand bender most of the time and reserve the hickey for when you need to create several really tight bends that a hand bender is unable to accomplish.

learn more about bender here

<u>brainly.com/question/9721733</u>

#SPJ4

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Scientists can help us understand the impact of human activities on the environment. True or false?
zhannawk [14.2K]

Answer:

True

Explanation:

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2 years ago
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res
damaskus [11]

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields  have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

3 0
3 years ago
One hypothesis states that plate movement results from convection currents in the
brilliants [131]
I believe is A) Inner core
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3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
A car accelerates uniformly in a straight line
erik [133]

Answer:

21.59 m/s

Explanation:

recall that one of the equations of motions can be expressed as

v² = u² + 2as

where,

v = final velocity (we are asked to find this)

u = initial velocity = 0m/s (because it says that it starts from rest)

a = acceleration = 3.7m/s²

s = distance travelled = 63 m

simply substitute the known values above into the equation:

v² = u² + 2as

v² = 0² + 2(3.7)(63)

v² = 466.2

v = √466.2

v = 21.59 m/s

3 0
3 years ago
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