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aev [14]
3 years ago
5

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

Physics
1 answer:
finlep [7]3 years ago
6 0
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.
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Every morning Mr. Jason travels to school 20 miles and it takes him to get there 30 minutes. How fast was he traveling?
ivanzaharov [21]
20miles-30mins
40miles/h
6 0
3 years ago
Read 2 more answers
Convertir:<br> A. 3Km a m<br> B. 250 ma Km<br> C. 1000Cm a m<br> D. 10000 mm a Cm
Katen [24]

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

4 0
3 years ago
What is the mass in kg of a leopard in a tree if the tree branch is 36 m up and the leopard's gravitational potential energy is
ElenaW [278]

Answer:

83.3kg

Explanation:

GPE = m × g × h

GPE = mass of leopard × 10 × 36m

29988J = 360 × mass

mass = 83.3kg

5 0
3 years ago
*WILL MARK BRAINLIEST*
Arte-miy333 [17]

Answer:

1.52

Explanation:

5 0
3 years ago
Read 2 more answers
An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period
KatRina [158]

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

   The period of the  asteroid is   T =  176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s

Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

7 0
2 years ago
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