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3 years ago

In which environment might you expect a deposit of rock salt or sock gypsum to have formed?

Physics

2 answers:

wolverine [178]3 years ago

8 0

Most likely in a shallow sea environment.

ArbitrLikvidat [17]3 years ago

5 0

Answer:

Option (B)

Explanation:

Rock salt is commonly known as the halite whose chemical formula is NaCl and gypsums are also soft minerals, forming large colored crystals and are commonly found to be present in association with the rock salt and sulfur deposits. They are also comprised of impurities such as sylvite and sometimes potassium is also found to be present in the form of a mineral. They are formed due to the evaporation of salt water that is mainly made up of dissolved ions namely Na+ and Cl-. They are found mainly in the shallow marginal seas, bays, and estuaries and in the dry lake beds.

Thus, the most likely place for the rock salt and gypsum to occur is the shallow sea.

Hence, the correct answer is option (B).

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Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar

den301095 [7]

Answer:

electric field Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

a) Let's write the electric field for each charge and the total field

E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

Et = E1 + E2 + E3

E1 = k q / (x-a)²

E2 = k (-2q) / x²

E3 = k q / (x + a)²

Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ where x << 1, for this we take factor like x from all the equations

Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

(1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

(1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0

3 years ago

Select all that apply.

GenaCL600 [577]

The correct answer to the question is: A) miles/hour and B) metre/ second.

EXPLANATION:

Before answering this question, first we have to understand speed.

The speed of a body is defined as the rate of distance travelled or the distance travelled by a body per unit time.

Hence, it is a derived quantity which is obtained from distance and time.

The unit of distance can be metre, miles, and the unit of time can be second, minutes or hour.

As speed is the distance covered per unit time, the perfect units will be miles/hour and metre/second.

Hence, the correct options are first and second.

5 0

3 years ago

Read 2 more answers

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$ .

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$ .

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0

3 years ago

How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?

maksim [4K]

Answer:

32000 N

Explanation:

Force Calculater

8 0

3 years ago

Other quanto

Alex73 [517]

I'm not sure what you were trying to put here

5 0

3 years ago