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4 years ago

Which electrode (anode or cathode is designated as positive in an electrolytic cell?

1 answer:

4 0

Electrodes are part of an electrochemical cell that uses chemical reaction to produce energy,called voltaic cell. The electrode is defined with its flow of current, and according to this it is called a anode or cathode. Anode is designated as positive and attracts electrons (anions). On the other hand, the cathode is designated as negative and attracts positive charge (cations).

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After the Collision the two car stick together find the final velocity of the two cars

adelina 88 [10]

Answer:

The two blocks collide in a totally inelastic collision (That means they stick together after they collide). What is their common final velocity after the inelastic collision? ... So we will find the total momentum initially, before the collision, and set that equal to the ... Example: Two cars collide at an intersection as sketched below.

Explanation:

8 0

4 years ago

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 2

Zolol [24]

Answer:

The radius is $r_1 = 0.315m$

The total charge is $Q= 2.912*10^{-8}C$

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

The magnitude of the electric field is $E= 2641.3 V/m$

The velocity is $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

The potential is $V_1 = 832 V$

The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

The potential at the radial distance is $V_2 = 499V$

The potential at the surface of the sphere is mathematically represented as

$V = \frac{kQ}{r}$

$Vr = kQ$

Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change

This means that

$V_1 r_1 = V_2 r_2$

Where $r_1$ is the radius of the sphere

and $r_2$ is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as

$r_2 = r_1 + d$

Substituting this into the equation

$v_1 r_1 = V_2 (r_1 +d)$

Now substituting value

$832 * r_1 = 499 * (r_1 + 0.21)$

$832r_1 - 499r_1 = 104.79$

$333r_1 = 104.79$

$r_1 = \frac{104.79}{333}$

$r_1 = 0.315m$

From the equation above

$V = \frac{kQ}{r_1}$

making Q the subject

$Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

Substituting into the equation

$Q =\frac{832 * 0.315}{9*10^9}$

$Q= 2.912*10^{-8}C$

According to Gauss law the electric field from outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

$E = \frac{kQ}{r_1^2}$

Substituting values

$E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

$E= 2641.3 V/m$

According the the law of energy conservation

The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

$EPE = V * e$

Where is e is an electron

And Kinetic energy is mathematically represented as

$KE = \frac{1}{2} m v^2$

From the statement above

$V_2 e = V_1 e + \frac{mv^2}{2}$

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

$V_2 e = \frac{mv^2}{2}$

The charge an electron has a value $e = 1.602*10^{-19}C$

And the mass of an electron is $m = 9.109 *10^{-31}kg$

Making v the subject

$v = \sqrt{\frac{2V_2 e}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }$

$v= 1.76 *10^{14} m/s$

7 0

3 years ago

When you drop a 0.43 kg apple, Earth exerts

N76 [4]

The magnitude of the Earth's acceleration is $7.0\cdot 10^{-25} m/s^2$

Explanation:

First of all, we start by calculating the magnitude of the force exerted by the apple on the Earth. According to Newton's third law, this is equal to the force exerted by the Earth on the apple, which is the weight of the apple, given by:

$F=mg$

where

m = 0.43 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F=(0.43)(9.8)=4.2 N$

Now we can find the Earth's acceleration by applying Newton's second law:

$F=Ma$

where:

F = 4.2 N is the net force exerted by the apple on the Earth

$M=5.98\cdot 10^{24} kg$ is the mass of the Earth

a is the Earth's acceleration

And solving for a, we find:

$a=\frac{F}{M}=\frac{4.2}{5.98\cdot 10^{24}}=7.0\cdot 10^{-25} m/s^2$

Learn more about Newton laws of motion:

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8 0

4 years ago

As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.01 m l

Valentin [98]

Answer:

The strength of the magnetic field is $75.6\ \mu T$ .

Explanation:

Given that,

Length of the rod, L = 1.01 m

Speed with which the rod is moving, v = 3.47 m/s

We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

$\epsilon=Blv$

B is magnetic field

$B=\dfrac{\epsilon}{lv}\\\\B=\dfrac{0.265\times 10^{-3}}{1.01\times 3.47}\\\\B=75.6\ \mu T$

So, the strength of the magnetic field is $75.6\ \mu T$ .

6 0

4 years ago

Study the scenario.

Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

7 0

3 years ago

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