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Goshia [24]
3 years ago
9

A jeweler needs to electroplate gold onto a bracelet using an ionic solution. He knows that the charge carriers in the ionic sol

ution are gold ions of charge e, and that each gold ion has a mass of 22g. The gold ions move through the solution, and are deposited on the bracelet. He has calculated that he must deposit 0.20 g of 901d to reach the necessary thickness. If each gold ion has a mass of g, what current should he use to electroplate the bracelet in three hours
Physics
1 answer:
Studentka2010 [4]3 years ago
7 0

Explanation:

First, we will calculate the total charge as follows.

      q_{total} = \frac{\text{weight of gold to be deposited}}{\text{atomic mass of gold}} \times N_{A} \times \text{charge on electron}

                    = \frac{0.2 g}{196.97 g/mol} \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19} C

                    = 97.80 C

Now, we will calculate the current required by the jeweler to plate the bracelet as follows.

                   i = \frac{q_{total}}{t_{total}}

                     = \frac{97.80 C}{3 \times 60 \times 60 sec}

                     = 9.05 \times 10^{-3} A

or,                  = 9.05 mA

thus, we can conclude that 9.05 mA current should he use to electroplate the bracelet in three hours.

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A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23 m/s. If the ball was caught 42 m from the
ExtremeBDS [4]

Answer:

a) v_{0y} = 9.72 m/s

b) v_{0x} = 20.85 m/s

c) t = 2.01 s

d) h_{max} = 4.82 m

Explanation:

a) The initial vertical velocity is given by:

v_{0y} = v*sin(\theta)

Where:

θ: 25°

v: is the magnitude of the speed = 23 m/s

v_{0y} = 23 m/s*sin(25) = 9.72 m/s

b) The initial horizontal velocity can be calculated as follows:

v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s

c) The flight time can be calculated using the following equation:

v_{0x} = \frac{x}{t}

Where:

x: is the total distance = 42 m

t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s

d) The maximum height is given by:

v_{fy}^{2} = v_{0y}^{2} - 2gh_{max}

Where:

v_{fy}: is the final vertical velocity =0 (at the maximum heigth)

g: is the gravity = 9.81 m/s²

h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m

I hope it helps you!                                  

7 0
3 years ago
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up
diamong [38]

Answer:

True, check attachment for code

Explanation:

To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.

The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.

The third line sets of a second String variable called result.

The fourth line is where the conversion is done.

We can compare the string

We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.

Check attachment for the code

3 0
3 years ago
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