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2 years ago

Principle of electroplating

Physics

2 answers:

frosja888 [35]2 years ago

4 0

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uysha [10]2 years ago

3 0

Answer:

Electroplating is the method of depositing one metal over another in the presence of a metal salt (in aqueous solution). The water molecule is released as the final product in this process. As a consequence, electroplating is based on the theory of hydrolysis.

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Plz help...Thanks.

Alexus [3.1K]

Answer: B) 33 amu

Explanation:

According to the table, the mass of a proton is 1.0073 amu and the mass of a neutron is 1.0087 amu.

To find the mass, we have to perform the following calculation:

15 x 1.0073 + 18 x 1.0087 = 33.2661 amu

Rounding up, the approximate mass of the nucleus is 33 amu.

7 0

3 years ago

The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle θ. One track

Elena L [17]

Answer:

a). $H=2.45m$

b). $H_{max}=1.94m$

Explanation:

For the block that stays on the track, its maximal height is attained when all of the kinetic energy is converted to potential energy

a).

The height for the block on the longer track can by find using this equation:

$\frac{1}{2}*m*v_o^2=m*g*H$

Cancel the mass as a factor in each element in the equation

$H=\frac{v_o^2}{2*g}$

$H=\frac{(6.94m/s)^2}{2*9.8m/s^2}$

$H=2.45m$

b).

The other lost some kinetic energy so, use a projectile motion to determine the total height for the other bock:

$E_k=E_p$

$E_k=m*g*H_1$

$E_k=\frac{1}{2}*m*v_o^2-\frac{1}{2}*m*v^2$

$m*g*H_1=\frac{1}{2}*m*(v_o^2-v^2)$

Solve to v'

$v^2=v_o^2-2*g*H_1$

$v=\sqrt{v_o^2-2*g*H_1}=\sqrt{(6.94m/s)^2-2*9.8m/s^2*1.25m}$

$v=4.8m/s$

$H_{max}=H_1+\frac{v^2*sin(50)}{2*g}=1.25m+\frac{(4.8m/s)^2*sin(50)}{2*9.8m/s^2}$

$H_{max}=1.94m$

8 0

3 years ago

Read 2 more answers

"it was so cold yesterday that the temperature only reached 275." which temperature scale is being used? what would be the corre

kow [346]

Remembering back to science I believe that temperature is on the Kelvin scale. The corresponding temperature on the Celsius scale would be very close to zero degrees which is the freezing point. The corresponding temperature on the Fahrenheit scale would be 32 degree which again is the freezing point.

7 0

2 years ago

Read 2 more answers

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

2 years ago

2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. Wh

Studentka2010 [4]

Answer:

2) deflection must be towards the negative side of the voltage.

4) the correct statements are: b and c

Explanation:

2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by

E = - N d $\phi_{B}$ / dt

where $\phi_{B}$ = B. A

The bold are vectors

Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.

4) From the Faraday equation we can see that the inductive electromotive force depends

* The magnitude of B that changes over time

* The area of the loop that changes over time

* The angle between B and the area that changes over time

* A combination of the above

With this analysis we will review the different alternatives given

a) False. It takes a temporary change and an absolute value of B

b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases

c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller

d) False. A temporary change of area is needed, in addition to increasing the area the current increases

We can see that the correct statements are: b and c

5 0

3 years ago

Principle of electroplating​

Principle of electroplating