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asambeis [7]
3 years ago
8

Problems with solar energy include _____.

Physics
2 answers:
german3 years ago
4 0
First choice:  the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low.  But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice:  the inability of current technology to store
captured solar energy

No.  We're pretty good at building batteries to store small amounts, or raising water to store large amounts.  Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice:  inconsistencies in the availability of the resource

I think this is it.  If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice:  lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.
Yuri [45]3 years ago
4 0

The answer is the first three.

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A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal
melamori03 [73]

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

\sum M__A }=0

- 40 (180) - (80) F_E Cos 37° + 55  

-7200 + F_E (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  F_E

- 30.79  F_E = 7200

F_E = -\frac{7200}{30.79}

F_E  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

\sum f_y = 0

F_E sin 37 + F_A cos 20 - F_G = 0

- 233.84 sin 37 + F_A cos 20 - 180 = 0

F_A cos 20  = 320.73

F_A = \frac{320.73}{cos 20}

F_A = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

\sum f_y = 0

F_E cos 37 - F_{CB} + F_A sin 20° = 0

-233.84 cos 37 - F_{CB} + 341.31 sin 20° = 0  

-186.75 -  F_{CB} + 116.74 = 0

-  F_{CB} -70.01 = 0

-  F_{CB} = 70.01

F_{CB} = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

7 0
4 years ago
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