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4 years ago

What kind of electric cell is considered to be rechargeable?

Physics

2 answers:

liubo4ka [24]4 years ago

8 0

Answer:

a. Storage cell

Explanation:

A Storage cell is an electric cell that can be recharged. During recharging of a Storage cell, current flow is reversed and oxidation occurs on the positive electrode and reduction on the negative electrode. Since the anode is by definition the electrode on which oxidation occurs and the cathode where reduction occurs, the positive electrode is now the anode and the negative is the cathode.

Phantasy [73]4 years ago

3 0

What kind of electric cell is considered to be rechargeable?
A. Storage cell

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mars1129 [50]

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

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3 years ago

What holds the moon in orbit around the earth?<br> A. The sun's gravity<br> B. The Earth's gravity

irina [24]

Answer:

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Explanation:

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3 years ago

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Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$