I tried to experiment of dropping candy into a liter of soda and caused a big explosion of foamy soda

What type of rock is limestone? Describe how a limestone rock is likely to change over a long period of time.

tatuchka [14]

Answer:

Limestone is a sedimentary rock

Explanation:

1. Limestone is a carbonate sedimentary rock made up of calcite(CaCO₃).This rock type forms calcareous shells and tests of organisms that was deposited in a sedimentary basin.

Limestone is used in building and construction works. It also finds application in chemical industries.

2. Over a long period of time, we would take a look at the rock "limestone" through the rock cycle.

Limestone being a sedimentary rock would be converted to marble, a metamorphic rock if subjected to metamorphic conditions over an extensive period of time. With series of metamorphic transformation, marble can grade to higher metamorphic facies of rocks as it combines with other minerals in the crust. The minerals would eventually change and as the changed rock approaches its melting temperature, melt would result.

From the other spectrum, limestone can be weathered if subjected extensively to denudation forces such as wind, water and glaciation. Water is more potent for the chemical weathering of limestone. Limestone would easily and readily dissolve in it over a long period of time.

7 0

3 years ago

If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would

7nadin3 [17]

Answer:

decrease by a factor 10

Explanation:

The parallax angle of a close star is given by

$p=\frac{1}{d}$

where

p is the parallax angle

d is the distance of the star from Earth, in parsecs

From the formula we see that the parallax angle is inversely proportional to the distance.

In this problem, the distance of the star is increased by a factor 10:

d' = 10 d

so the new parallax angle would be

$p'=\frac{1}{10 d}=\frac{1}{10}\frac{1}{d}=\frac{p}{10}$

So, the parallax angle would decrease by a factor 10.

4 0

3 years ago

In an inertia balance, a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the ac

avanturin [10]

Answer:

2) f = 0.707 Hz

Explanation:

Given m₁ = 1.0 kg , f₁ = 1.0 Hz

So using the equation

f₁ = ( 1 / 2 π ) * √K / m₁

Solve to determine K' constant of spring

K = m * ( 4 π ² * f ² )

K = 1.0 kg * ( 4 π ² 1.0² Hz )

K = 39.4784176

So given 2.0 kg the frequency can be find using formula

f₂ = ( 1 / 2 π ) * √K / m₂

f₂ = ( 1 / 2 π ) * √39.4784176 / 2.0 kg

f₂ = 0.707 Hz

4 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

What are the components of vector b

Afina-wow [57]

Answer:

G

Explanation:

G

5 0

3 years ago