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2 years ago

The average salinity of sea water is A. 3% B. 3.5% C. 2.5% D. 2%

Physics

2 answers:

mel-nik [20]2 years ago

8 0

The average salinity of seawater is approximately 3.5%, or 35 per thousand.

nadezda [96]2 years ago

7 0

Answer:

In terms of percent, the salinity of most sea water is between 2.5 and 4.5.

Explanation:

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

2 years ago

All living things are made up of two or more cells true or false

Virty [35]

The answer is false because there are single celled organisms.

5 0

3 years ago

In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the dist

weeeeeb [17]

Answer: The multiplication factor is 72.136 cm. This will give you the unit conversion when multiplied with 28.4 inch

Explanation:

1 inch = 2.54 cm

28.4 inches = x cm

Xcm= (28.4 inches × 2.54cm)/1 inch

X= 72.136

6 0

3 years ago

Suppose you drive an average of 25 miles a day, for a total of 60 years in a lifetime. The total distance you would have covered

Ahat [919]

The total distance you would have covered is equivalent to going around the earth 22 times.

<h3>Total distance traveled</h3>

The total distance traveled at a given speed and time of motion is calculated as follows;

Distance = speed x time

$Distance = \frac{25 \ miles}{day} \times \frac{365 \ days}{1 \ year} \times 60 \ years\\\\Distance = 547, 500 \ miles$

<h3>Distance round the Earth</h3>

The distance round the earth or circumference of the earth of the earth has been given as 25,000 miles

<h3>Number of times round the earth</h3>

$n = \frac{547,500}{25,000} \\\\ n = 21.9 \ times \ \approx 22 \ times$

Thus, the total distance you would have covered is equivalent to going around the earth 22 times.

Learn more about distance here: brainly.com/question/17273444

3 0

2 years ago

Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced fro

soldier1979 [14.2K]

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of copper wire,d=0.0113 in

Radius of copper wire= $r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in$

$\frac{1}{1000}=10^{-3}$

Density of copper= $\rho=8.96g/cm^3$

1 lb=454 g

3.25 lb= $3.25\times 454=1475.5 g$

Mass of Azurite= $1475.5 g$

Mass of copper= $\frac{55.1}{100}\times 1475.5=813 g$

Density= $\frac{Mass}{volume}$

Using the formula

$8.96=\frac{813}{volume\;of\;copper}$

Volume of copper wire= $\frac{813}{8.96}=90.7cm^3$

Radius of copper wire= $5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm$

1 in=2.54 cm

Volume of copper wire= $\pi r^2 h$

$\pi=3.14$

Using the formula

$90.7=3.14\times (14.35\times 10^{-3})^2\times h$

$h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}$

$h=140273 cm$

1 m=100 cm

$h=\frac{140273}{100}=1402.73 m$

Hence, the length of copper wire required=1402.73 m

7 0

3 years ago