Matter in which particles are arranged in repeating geometric patterns is a _____

Gekata [30.6K]

Answer: is option C: <em>Prolonged periods of cooling and warming</em>.

Explanation:

In the history of Earth, climate varies time to time. At times, the Earth's atmosphere was much hotter and humid as compare to the present time, but similarly it has been noticed that climate also has been much colder than he present time, whereas the number of glaciers covers much of the Earth's surface. There are two kinds of periods in which we further classified Earth's climate namely, Glacial period, and Inter-glacial periods. It has been noticed that the average global temperature of Earth during glacial periods was around 5.5°C or 10°F, which is less than Earth's present climate. On the other hand, during inter-glacial periods Earth's temperature was about 1.1°C or 2.0°F, which is again higher as compared to current temperature. Over the past 900,000 years, Earth's temperature varied less than 5°C. Scientist believe by looking at the Earth's climate history, that glaciers will proceed again in formation, but it will take thousands of years.

8 0

3 years ago

Please help.<br>Just True or False

True [87]

Answer:

1. T

2. F

3.F

4.T

thank you

6 0

3 years ago

Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p

Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

$\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v$

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = $\frac{2\times r^2(d-\rho)}{9\times n}$

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = $[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}$ [/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= $\frac{100}{387\times10^{-5}}$

= 2.6 x 10⁴ s.

5 0

3 years ago

One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0