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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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Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at

miv72 [106K]

Answer:

The angle at which the boat must head is $- 22.47^{\circ}$

Solution:

As per the solution:

Distance between the parallel banks, d = 40 m

The maximum speed of water, v' = 3 m/s

constant speed, u' = 5 m/s

Also,

The speed of water of the river at a distance of 'x' units from the west bank is given as a sine function:

$f(x) = 3sin(\frac{\pi x}{40})$ (2)

Now, to determine the angel at which the boat must head:

The velocity of the engine of the boat:

v = $u'cos\theta\hat{i} + u'sin\theta\hat{j}$

v = $5cos\theta \hat{i} + 5sin\theta\hat{j}$

The abscissa of the boat at time t:

v = $5cos\theta t\hat{i}$

Now, from above and eqn(1) , we can write:

$f(5cos\theta t) = 3sin(\frac{\pi \times 5cos\theta t}{40})$

Now, boat's velocity at time t:

v = $5cos\theta \hat{i} + (5sin\theta + 3sin(\frac{\pi \times 5cos\theta t}{40})\hat{j}$

In order to obtain the position of the boat, we integrate both the sides, we get:

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C (3)

Now, at r = 0:

0 = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C

C = $\frac{24}{\pi cos\theta}\hat{j}$

Now, from eqn (3)

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40} + \frac{24}{\pi cos\theta})\hat{j}$ (4)

the baot will reach the point at y = 0 and x = 40

Now,

40 = $5cos\theta t$

$t = \frac{8}{cos\theta}$

Substituting the above value of 't' in eqn (4):

r = $5sin\theta \frac{8}{cos\theta}\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta \frac{8}{cos\theta}}{40} + \frac{24}{\pi cos\theta})\hat{j}$