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3 years ago

Chemical rock formed by crystallization of excess dissolved minerals

Physics

2 answers:

Alexandra [31]3 years ago

8 0

Answer: Chemical sedimentary rocks.

Explanation: Those may be a type of sedimentary rocks called "chemical sedimentary rocks". This type of rock forms when mineral constituents in solution become supersaturated and inorganically precipitate. Formed by deposit of dissolved substances, were the largest volume corresponds to masses of salts accumulated by supersaturation of seawater. When seawater is stagnant, it begins to evaporate and dissolved minerals precipitate, this process gives rise to evaporites, for example gypsum and gem salt.

amm18123 years ago

7 0

Answer:

heyooo!!!

Sedimentary

hope this helps!!

Explanation:

When the sediments harden, the layers are preserved. Sedimentary rocks formed by the crystallization of chemical precipitates are called chemical sedimentary rocks.

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Two positive point charges are 4.9cm apart. If the electric potential energy is 70.0 μJ, what is the magnitude of the force betw

ExtremeBDS [4]

Hi there!

Recall the following:

$V \text{ (Electric Potential Energy) } = \frac{kq_1q_2}{r}\\\\F_E = \frac{kq_1q_2}{r^2}$

k = Coulomb's Constant (Jm/C²)

q = Charge (C)
r = distance between charges (m)

To calculate the electric force between the two charges, we can simply divide by another 'r' (distance):

$F_E = \frac{70}{0.049} = \boxed{1428.57 \mu J}$

6 0

2 years ago

A boy rubs two balloons against his sweater. One balloon acquires a charge of 3.0 E -6 C. The other balloon acquires a charge of

KonstantinChe [14]

Answer:

F= 67.5 N

Explanation:

We use the equation for the Coulomb's Law of Force between two charges Q1 and Q2 (in Coulombs) separated by a distance d (in meters):

$F=k\frac{Q1*Q2}{d^2}$

where the constant k is the Coulomb's constant ( $9*10^9 N\frac{m^2}{C^2}$

The charges are $Q1=3.0*10^{-6}C$ , Q2=2.5*10^{-7}C; the distance we convert into meters to match the appropriate units of the Coulomb constant k (1 cm = 0.01 m)

Now we input all these data into the equation, knowing that given the appropriate units, the force will be expressed in Newtons (N):

$F=k\frac{Q1*Q2}{d^2}\\F=9*10^9\frac{3.0*10^{-6}*2.5*10^{-7}}{0.01^2} N\\F= 67.5N$

4 0

3 years ago

Which phrases describe all the outer planets’ motion? Check all that apply.

uranmaximum [27]

Answer: slow revolution and fast rotation

Solar system has 8 planets. 4 inner rocky planets - Mercury, Venus, Earth and Mars and 4 outer gaseous planets - <u>Jupiter, Saturn, Uranus and Neptune.</u> The outer planets have few common features.

They are gaseous. There period of revolution is larger than the inner planets which means that they have slow revolution about the Sun. One day on the outer planets is smaller than the inner planets which means they have fast rotation.

<u>For example,</u> Jupiter has revolves around sun in 11.86 Earth years and rotates about axis in 9.8 Earth hours. Uranus revolves around sun in 84 Earth years and rotates on its axis 17.9 Earth hours.

8 0

3 years ago

Read 2 more answers

Define 1 unit electricity

Tomtit [17]

Answer:

A unit is represented in kWH or Kilowatt Hour. This is the actual electricity or energy used. If you use 1000 Watts or 1 Kilowatt of power for 1 hour then you consume 1 unit or 1 Kilowatt-Hour (kWh) of electricity.

4 0

3 years ago

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

3 years ago