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elena-s [515]
4 years ago
14

Which of the following statements about ocean water is correct?

Physics
2 answers:
shusha [124]4 years ago
8 0
A would be true. the volder, the more dense
Bas_tet [7]4 years ago
3 0

Sea water contains 3.5% salts, dissolved gasses, organic substances and undissolved particulate matter. The presence of salts influences most physical properties of sea water to some degree but does not determine them. Some properties are not significantly affected by salinity. Two properties which are determined by the amount of salt in the sea are conductivity and osmotic pressure.

Ideally, salinity should be the sum of all dissolved salts in grams per kilogram of sea water. In practice, this is difficult to measure. The observation that - no matter how much salt is in the sea - the various components contribute in a fixed ratio, helps overcome the difficulty. It allows determination of salt content through the measurement of a substitution quantity and calculation of the total of all material making up the salinity from that measurement.

Determination of salinity could thus be made through its most important component, chloride. Chloride content was defined in 1902 as the total amount in grams of chlorine ions contained in one kilogram of sea water if all the halogens are replaced by chlorides. The definition reflects the chemical titration process for the determination of chloride content and is still of importance when dealing with historical data.

Salinity was defined in 1902 as the total amount in grams of dissolved substances contained in one kilogram of sea water if all carbonates are converted into oxides, all bromides and iodides into chlorides, and all organic substances oxidized. The relationship between salinity and chloride was determined through a series of fundamental laboratory measurements based on sea water samples from all regions of the world ocean and was given as

So the correct answer would be

High salinity and cold temperature result in denser water.

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In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff
Fynjy0 [20]

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

8 0
3 years ago
Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick
77julia77 [94]

Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)

Solids' molecules are strongly attracted to one another. As a result, the molecules are barely moving and tightly packed. Because of this, shape and volume are fixed.

The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).

You can learn more about states of the matter here:

brainly.com/question/18538345

#SPJ4

6 0
2 years ago
Bernice draws an oxygen atom. She draws a small circle for the nucleus. Inside of the circle, she draws plus signs for protons a
Lelu [443]

Answer

It should be A and C

Explanation:

because oxygen is number 8 in the periodic table of elements and has a atomic weight of 15.999 you use those numbers to figure out what is true between those.

The 8 for oxygen goes for the number of electrons and proton and to find neutrons u round the 15.999 up which now make it 16 and subtract it by the 8 now you have 8 protons, 8 neutrons, and 8 electrons

5 0
3 years ago
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle
marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

n_1sin\theta_1 = n_2sin\theta_2

Where,

\theta = Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

n_1 sin\theta_1 = n_2sin\theta_2

(1.54) sin\theta_1 = (1.33)sin(90)

sin\theta_1 = \frac{1.33}{1.54}

\theta = sin^{-1}(\frac{1.33}{1.54})

\theta = 59.72\°

Therefore the \alpha_{max} would be equal to

\alpha = 90\°-\theta

\alpha = 90-59.72

\alpha = 30.27\°

Therefore the largest value of the angle α is 30.27°

3 0
3 years ago
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail
Alik [6]

Answer:

1.34\cdot 10^{-16} C

Explanation:

The strength of the electric field produced by a charge Q is given by

E=k\frac{Q}{r^2}

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C

4 0
3 years ago
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