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Rasek [7]
3 years ago
9

What are some goals of NYFEA? Select three options.

Engineering
2 answers:
shutvik [7]3 years ago
8 0

Answer:

developing agricultural leaders

encouraging new and young farmers

encouraging small farms

Explanation:

got it correct

Brianna bts
2 years ago
edge let me guess
faltersainse [42]3 years ago
7 0

Answer:

developing agricultural leaders

encouraging new and young farmers

encouraging small farms

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KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown
kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

i_x+50mA=i_y\\i_x-i_y=0.05A

Also

V_{\Delta}=1K*i_y

Now applying KVL on loop 1 as indicated in the attached figure

1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0

Similarly for loop 2

2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0

So the system of equations become

i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0
3 years ago
6. Driving with parking lights only (in place of headlights) is against the law. A. True B. False
trasher [3.6K]

Answer:

B false it is illegal to only have got fog lights on though and bright headlights because it can distract other drivers going last and if the y are distracted then that will cause a collision

Hope this helps :)

Explanation:

4 0
3 years ago
Read 2 more answers
Isormophous phase diagram
shusha [124]

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0
3 years ago
Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th
sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, \sigma is given by

\sigma=\frac {F}{A} where F is force and A is area and area is given by \frac {\pi d^{2}}{4} hence

\sigma=\frac {4F}{\pi d^{2}} where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}

\sigma \approx 127 Mpa

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

\epsilon=\frac {\triangle l}{l} where\epsilon is the strain, \triangle l is elongation and l is original length and making elongation the subject

\triangle l= \epsilon \times l and substituting strain with 0.0015 and length l with 100.6 mm then

\triangle l=0.0015\times 100.6=0.1509 mm

(b)

Lateral strain is given by

\epsilon_{lat}=\frac {\triangle d}{d} and substituting -v\epsilon for \epsilon_{lat} where v is poisson ratio then

-v\epsilon=\frac {\triangle d}{d} and making \triangle d the subject then

\triangle d=-vd\epsilon and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

\triangle d=-(0.35)*10*0.0015=-0.00525 mm and the negative sign indicates decrease in diameter

8 0
3 years ago
when a metal, such as lead, is oxidied (loses electrons) to form a positive ion (cation), how does he solubility change?
o-na [289]

Answer: The size of the ion and the charge of the ion are the factors that affect solubility in water.

Explanation:

Lead lose electrons to become cations. Compounds with small ions tend to be less soluble than compounds with large ions. Large ions have higher solubility. This is because small ions are closely packed so it is difficult for water to break them apart.

Compounds with small ions seemingly have less solubility than those with large ions. The ions in the compound attract each other, and the water molecules attract the ions. Compounds would be soluble in water If the water molecules have a greater or higher attraction to the ions than ions have for each other.

8 0
4 years ago
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