What is the solution set of the equation 3|5 -x| + 2 = 29?

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

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

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

PLease Help!

scoundrel [369]

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

Slope of a graph = Change in value of Y -axis / Change in values of X -axis

Comparing both the equations

Change in value of Y -axis = Change in velocity in velocity

Change in values of X -axis = Time taken

So velocity values should be on the Y axis and Time values should be on the X axis.

Option A is the correct answer.

8 0

3 years ago

You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje

Over [174]

Answer: a. m = 7.7 kg

b. V = 435.52 in³

c. m = 1927 kg

d. V = 335.37 cm³

e. m = 3 kg

Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

$\rho=\frac{m}{V}$

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

$\rho=\frac{m}{V}$

$m=\rho.V$

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

$m=\rho.V$

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as $V=\frac{4}{3}.\pi.r^{3}$

Diameter is twice the radius, then r = 4.31 cm.

Volume is

$V=\frac{4}{3}.\pi.(4.31)^{3}$

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. $m=\rho.V$

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0

3 years ago

0.

larisa [96]

Answer:21

Explanation:

4 0

2 years ago

Who performed classic experiments that supported the semiconservative model of dna replication?

Salsk061 [2.6K]

Meselson and Stahl

Explanation:

The classic experiment that supported the semiconservative model of dna replication was performed by Matthew Meselson and Franklin W. Stahl. In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. They used E. coli bacteria as a model system.

6 0

3 years ago