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OLEGan [10]
2 years ago
8

Many farms and ranches use electric fences to keep animals from getting into or out of specific pastures. When switched on, an e

lectric current is produced in the fence. When an animal touches the electrified fence, it receives a small shock. What material would be the best choice for making an effective electric fence, and why?
Engineering
2 answers:
Nikolay [14]2 years ago
7 0

Answer:

Aluminum

Explanation:

The best material to use when creating an electric fence would be Aluminum. Aluminum wiring is incredibly durable and can be easily obtained. Since aluminum is a non-magnetic metal its conducting capabilities far exceed other metallic options in the market and is also why companies choose aluminum for their high tension cable wiring. Aside from being more expensive than other feasible options its durability and conducting capabilities make it easily the best option.

madreJ [45]2 years ago
3 0

Answer:

Steel Wires

Explanation:

International house of pancakes

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The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%
Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

4 0
3 years ago
Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di
blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0
3 years ago
11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h
IrinaVladis [17]

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

7 0
3 years ago
The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter
timurjin [86]

Answer:

It’s called Expansion Joints

6 0
2 years ago
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0
3 years ago
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