All categories

3 years ago

Now put the names of the planets in increasing order based on their distance from the sun

1 answer:

5 0

The eight planets of the Solar System arranged in order from the sun:

Mercury: 46 million km / 29 million miles (.307 AU)
Venus: 107 million km / 66 million miles (.718 AU)
Earth: 147 million km / 91 million miles (.98 AU)
Mars: 205 million km / 127 million miles (1.38 AU)
Jupiter: 741 million km /460 million miles (4.95 AU)
Saturn: 1.35 billion km / 839 million miles (9.05 AU)
Uranus: 2.75 billion km / 1.71 billion miles (18.4 AU)
Neptune: 4.45 billion km / 2.77 billion miles (29.8 AU)

Astronomers often use a term called astronomical unit (AU) to represent the distance from the Earth to the Sun.

+ Pluto (Dwarf Planet): 4.44 billion km / 2.76 billion miles (29.7 AU)

You might be interested in

Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set

VikaD [51]

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

$V = IR \rightarrow R = \frac{V}{I}$

Our values are

$I = 500mA = 0.5A$

$V = 37V$

Replacing,

$R = \frac{V}{I}$

$R = \frac{37}{0.5}$

$R = 74 \Omega$

Therefore the smallest resistance you can measure is $74 \Omega$

8 0

3 years ago

Why do people cover their flowers on a cool night. Think about radiation absorption conduction and temperature

Blababa [14]

When the temperature lowers the plants can freeze and die. One way to prevent that is to cover the plants so they dont freeze

3 0

3 years ago

Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t

ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body Mass (gram) Engine Force (newtons)

1 500 1 25

2 1500 2 20

3 750 3 30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

20 = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

$=\frac{200}{5}$

$= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $40 \ m/s^2$ at the start of the launch.

8 0

3 years ago

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$