All categories

3 years ago

Solve this code. Use the Periodic Table if necessary:

Physics

1 answer:

Leni [432]3 years ago

6 0

oxygen,
hydrogen,
sulfur,
iodine,
iron,
carbon,
nitrogen,
phosphorus,
calcium.

You might be interested in

PLEASE HELP THE question on picture

Gnoma [55]

Vertical:

(20 m/s) sin(25º) ≈ 8.45 m/s

Horizontal:

(20 m/s) cos(25º) ≈ 18.1 m/s

7 0

3 years ago

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?

$3.081x10^{16}m$

$9.46x10^{15}m$

(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

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumpe

Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

$U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where we have

$k=2\cdot 10^6 N/m$ is the spring constant of the bumper

x is the maximum compression of the bumper

$m=30\cdot 10^4 kg$ is the mass of the car

$v=2.0 m/s$ is the speed of the car

Solving for x, we find the maximum compression of the spring:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m$

8 0

3 years ago

Read 2 more answers

A car is moving at 76 miles per hour. the kinetic energy of that car is 5 × 105 j. how much energy does the same car have when i

valentina_108 [34]

Its would be 144. cause if u think about it

3 0

3 years ago

What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)

mariarad [96]

Answer:

The correct equation for measuring the average microscopic weight for 3 isotopes is multiply the rate of abundance by each weight and add them.

Explanation:

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

Take the correct weight of each isotope (that will be in decimal form)

Multiply the weight of each isotope by its abundance

Add each of the results together.

This gives the required average microscopic weight of the three isotopes.

3 0

3 years ago