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masya89 [10]
3 years ago
11

To increase the amount of electricity produced in the U.S., we could place what on top of all Wal-Mart stores?

Physics
2 answers:
Jlenok [28]3 years ago
4 0
The question will be looking for Solar Panels, as they are flat, non-intrusive, and reasonably low maintenance ie the most sensible choice.

However I mean in theory you <em>could</em> stick windmills on top of big Wal-Mart stores as wind turbines are placed on roofs regularly, so you can tell whoever set you that question that it is vague ;)
Alecsey [184]3 years ago
3 0

That's a great idea !  A wind turbine (windmill) or solar panels
on the roof of every WalMart, but don't stop there.  Let K-Mart,
Target, Sears, Macy's, 7-11, McDonald's, and Subway in on it too. 

In fact, why not place them on the roof of every house and
apartment building too ?

Just one question:  Who's going to pay for them ?
_________________________________________________

The biggest single year-round residential use of energy in the
USA is to heat up and keep a ready supply of hot water.  So the
single biggest residential use of energy could be eliminated with
a simple solar water heater on the roof of every house.  From my
own personal observation, I know that Israel is there now, but here
in the USA, it's apparently not worth the time, effort, or expense ! ? !

You might be interested in
k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s
Irina18 [472]

Answer:

T_ww = 43,23°C

Explanation:

To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

E_in=E_out+E_loss

The energy associated to a current of fluid can be defined as:

E=m*C_p*T_f

So, applying the energy balance to the system described:

m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss

Replacing the values given on the statement, we have:

1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8  kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30  kJ/s

Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

T_WW=43,23 °C

Have a nice day! :D

6 0
3 years ago
The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified th
hram777 [196]

Answer:

e) indicated that the speed of light is the same in all inertial reference frames.

Explanation:

In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium <em>aether</em>. They believed that even the space is not empty and filled with aether.

Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.

It thus proved that the speed of light remains same in all inertial frames.

Also, it became a base for the special theory of relativity by Einstein.

5 0
3 years ago
Based on the replacement reaction, what would the products of the reaction be?
cricket20 [7]

Answer:

\rm Be(OH)_2 and \rm (NH_4)_2 SO_4. The missing ion would be \rm OH^{-}.

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

  • \rm BeSO_4, and
  • \rm NH_4 OH.

In particular,

  • \rm BeSO_4 is made up of \rm Be^{2+} ions and \rm {SO_4}^{2-} ions, while
  • \rm NH_4 OH is made up of \rm {NH_4}^{+} ions and \rm OH^{-} ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

  • \rm Be^{2+} is a cation. In \rm BeSO_4, \rm Be^{2+} was bounded \rm {SO_4}^{2-} anions. During the reaction, it bonds with \rm OH^{-} anions to produce \rm Be(OH)_2.
  • \rm {NH_4}^{+} is also a cation. In \rm NH_4 OH, \rm {NH_4}^{+} was bounded to \rm OH^{-} ions. During the reaction, it bonds with \rm {SO_4}^{2-} anions to produce \rm (NH_4)_2 SO_4.

Hence, the two products will be \rm Be(OH)_2 and \rm (NH_4)_2 SO_4.

Note that charges on the ions must balance. For example, a \rm Be^{2+} ion carries twice as much charge as an \rm {NH_4}^{+} ion. As a result, each \rm Be^{2+} ion would bond with twice as many \rm OH^{-} ions as \rm {NH_4}^{+} would in \rm NH_4 OH.

4 0
3 years ago
Read 2 more answers
Term meaning "wise human"
Novay_Z [31]
'Intelligent person' could be a possibility for your answer, also a 'scientist' or a 'philosopher' as well as 'an old person' may equal to the meaning of being 'wise humans'.
7 0
3 years ago
Read 2 more answers
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
3 years ago
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