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3 years ago

11

Which transformation could take place at the anode of an electrochemical cell?

1 answer:

erica [24]3 years ago

7 0

As I found out the choices for your question which are:

<span>A) F2 to F-
B) Cr2O7²- → Cr2+
C) O2 to H2O
D) HAsO2 to As
</span>
Unfortunately, the answer does not belong to the choices provided. In fact, it is the oxidation half-reaction that occurs at the anode of an electrode for it to transform chemical energy to consumable electrical energy.

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3 years ago

Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5

dsp73

Answer:

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi\times 10^{-7}\ N/A^2$

$i_1$ = Current in first wire = 2.9 A

$i_2$ = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

7 0

2 years ago

calculate the load placed 10m from the fulcrum that can be balanced by an effort of 5 N applied at a distance of 4 m from the fu

Elena L [17]

Answer:

A = 2 m from fulcrum

Explanation:

Product of anti clockwise = Product of clockwise moment

5 × 4 = 10 × A

20 = 10 x A

A = 20 / 10

A = 2 m from fulcrum

8 0

2 years ago