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2 years ago

give an example of how the law of inertia is demonstrated (a) for moving objects and (b) for objects at rest

Physics

1 answer:

lakkis [162]2 years ago

6 0

Answer:

Both

Explanation:

Both of them are part of newtons laws of motion

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65. If Jill was traveling at 300 miles in 4 hours due South what was his velocity in miles per second,

iris [78.8K]

Correct answer is A. Divide 300 by 75

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2 years ago

Chose a substance your familiar with. What are it’s physical and chemical properties

finlep [7]

So, physical properties are what we can detect with our basic 5 senses or measuring tools, and the things that, when changed, dont actually change the chemical properties (like atoms and molecules). Lets take wood for an example: its brown, its solid, it can be big or small, it has a taste and smell, its boiling, freezing or melting point...

Chemical properties, on the other hand, are the things we can change with, for example, experiments and tools. Does it burn? Can it rust/oxidize? How does it react with other chemicals? Is it radioactive, or toxic? All of these are chemical properties you can probably answer.

6 0

3 years ago

2 (a) What is the distance from the Sun to Earth in terms of solar radii? Earth radii?

ohaa [14]

a) Distance Earth-Sun is 215.5 solar radii and 23,548 Earth radii

Mercury: 1.7 days

Mars: 6.5 days

Jupiter: 21.7 days

Uranus: 86.8 days

Neptune: 130.2 days

c) $1.3\cdot 10^6$ Earths can fit inside the Sun

Explanation:

The distance between the Sun and the Earth is 150 millions km, so

$d=150\cdot 10^6 km = 1.50\cdot 10^{11} m$

The solar radius is

$r_s = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m$

Therefore the distance Earth-Sun in solar radii is

$d_s = \frac{d}{r_s}=\frac{1.50\cdot 10^{11}}{6.96\cdot 10^8}=215.5$

The Earth radius is

$r_e = 6.37\cdot 10^6 m$

Therefore the distance Earth-Sun in Earth radii is

$d_e=\frac{d}{r_e}=\frac{1.50\cdot 10^{11}}{6.37\cdot 10^6}=23,548$

The speed of the solar wind is

$v=400 km/s = 4\cdot 10^5 m/s$

The value of 1 AU (Astronomical Unit) is

$1 AU = 1.50\cdot 10^{11}m$ (distance Earth-Sun)

The distance between the Sun and Mercury is:

$d=0.4 AU \cdot 1.50\cdot 10^{11}=6.0\cdot 10^{10} m$

So the time taken by a parcel of solar wind to reach Mercury is:

$t=\frac{d}{v}=\frac{6.0\cdot 10^{10}}{4.0\cdot 10^5}=150,000 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{150,000}{86400}=1.7 d$

The distance between the Sun and Mars is:

$d=1.5 AU \cdot 1.50\cdot 10^{11}=2.25\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Mars is:

$t=\frac{d}{v}=\frac{2.25\cdot 10^{11}}{4.0\cdot 10^5}=562,500 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{562,500}{86400}=6.5 d$

The distance between the Sun and Jupiter is:

$d=5 AU \cdot 1.50\cdot 10^{11}=7.5\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Jupiter is:

$t=\frac{d}{v}=\frac{7.5\cdot 10^{11}}{4.0\cdot 10^5}=1.88 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{1.88\cdot 10^6}{86400}=21.7 d$

The distance between the Sun and Uranus is:

$d=20 AU \cdot 1.50\cdot 10^{11}=3.0\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Uranus is:

$t=\frac{d}{v}=\frac{3.0\cdot 10^{12}}{4.0\cdot 10^5}=7.5 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{7.5\cdot 10^6}{86400}=86.8 d$

The distance between the Sun and Neptune is:

$d=30 AU \cdot 1.50\cdot 10^{11}=4.5\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Neptune is:

$t=\frac{d}{v}=\frac{4.5\cdot 10^{12}}{4.0\cdot 10^5}=11.3 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{11.3\cdot 10^6}{86400}=130.2 d$

As we said in part a), we have:

$r_e = 6.37\cdot 10^6 m$ (radius of the Earth)

$r_s=6.96\cdot 10^8 m$ (radius of the Sun)

So the volume of the Earth can be calculated as:

$V_e=\frac{4}{3}\pi r_e^3 = \frac{4}{3}\pi (6.37\cdot 10^6)^3=1.08\cdot 10^{21} m^3$

While the volume of the Sun is

$V_s=\frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (6.96\cdot 10^8)^3=1.41\cdot 10^{27} m^3$

Therefore, the number of Earths that could fit inside the Sun is:

$\frac{V_s}{V_e}=\frac{1.41\cdot 10^{27}}{1.08\cdot 10^{21}}=1.3\cdot 10^6$

Learn more about the Solar System:

brainly.com/question/2887352

brainly.com/question/10934170

#LearnwithBrainly

6 0

3 years ago

In your daily life you come across a range of motion in which acceleration is in the direction of motion

irakobra [83]

That makes no sense to me somehow

3 0

3 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

Given the following data:

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

Conversion:

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes