Who wants to be my freind

The texture of obsidian is best described as glassy.Why does obsidian contain few or no mineral grains? by the way I put it unde

Brrunno [24]

hi, so the material that would normally form mineral crystals does not have time to form a crystalline structure because of very rapid cooling after volcanically erupting into the air or onto the surface. Obsidian is a volcanic glass.

4 0

3 years ago

Explain how a star's absorption spectrum and the concept of red shift help support the Big Bang Theory.

marta [7]

Answer:

Red shift supports the big bang theory. ... The light from distant galaxies is red shifted (this tells us the galaxies are moving away from us) and the further away the galaxy the greater the red shift (this tells us that the more distant the galaxy the faster it is moving). Constellations look like they are moving because earth is rotating on it's axis.

May I have brainliest, please?

4 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago

Describe the evolution of a pulsar over time, in particular how the rotation and pulse signal changes over time.

madam [21]

Answer:

As beams of particles and their associated energy are given off, the pulsar will lose energy slowly, which will decrease the rate of its rotation. The frequency of pulses would therefore decrease, so that fewer pulses are observed in a given time span. The strength of the pulse signal will also decrease so the pulses will become fainter. Eventually, the pulsar should rotate so slowly and have such a low emission of radiation that it would no longer be observable.

3 0

4 years ago

Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

myrzilka [38]

Given: Mass of earth Me = 5.98 x 10²⁴ Kg

Radius of earth r = 6.37 x 10⁶ m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r

but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²

GMe = 4π²r³/T²

T² = 4π²r³/GMe

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds or around 1.4 Hour

3 0

3 years ago