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ryzh [129]
2 years ago
9

What’s the first step you would take to build a voltaic cell?

Physics
1 answer:
TiliK225 [7]2 years ago
3 0

Answer:

In a voltaic cell, current is produced by connecting an oxidation reaction half cell to a reduction reaction half cell in an electrolyte solution. Oxidation and reduction reactions (redox reactions) are chemical reactions involving a transfer of electrons from one atom in the reaction to another. When two different oxidation or reduction reactions are connected electrically by connecting the cathode to the anode, a current is formed. The direction depends on the type of reaction taking place at the terminal.

The first step would be to determine metals to be used as the cathode and the anode.

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Does current in Helmholtz coils flow in the same or opposite direction through each coil? Explain.
zhenek [66]
To get a uniform field in the central region between the coils, current flows in the same direction in each. 
5 0
3 years ago
1.10) For a fixed mass of gas at constant temperature, if the volume of the gas (V) increases to twice its original amount, the
Aliun [14]
B. 1/2p
From the relation
P1V1=P2V2
P2=P1V1/V2
If everything equals one
P2=1*1/2(1)
That would equal 1/2
I hope this helps
Please mark as brainliest
7 0
3 years ago
Read 2 more answers
Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from
qwelly [4]
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0
3 years ago
At which point is the kinetic energy the greatest?
mr_godi [17]

Answer:

I don't know

Explanation:

I think it's at the highest, learned this in seventh grade

6 0
3 years ago
Read 2 more answers
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
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