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3 years ago

In electroplating, the object to be plated is which part of an electrolytic cell? cathode or anode

Physics

2 answers:

Bingel [31]3 years ago

7 0

Since the anode dissolves and goes to the cathode. The object to be electroplated is kept at the cathode.

Please mark me as brainliest.

ValentinkaMS [17]3 years ago

5 0

The cathode would be the one to be plated and the anode would be either a sacrificial anode or an inert anode.

Good Luck!

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4 years ago

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Two identical blocks of iron are placed in contact, one at 10°C and the other at 20°C. If the cooler block cools to 5°C and the

Alenkinab [10]

Answer:

this is violation of II law of thermodynamics

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So here it says that initially two blocks are at 10 degree C and 20 degree C

Now the cool object will become cooler and hot object becomes hotter

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3 years ago

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

Change of Momentum

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$