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sergeinik [125]
3 years ago
9

Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What

are some physical and chemical changes it can undergo?
Physics
1 answer:
Gekata [30.6K]3 years ago
6 0
<span>you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)

</span><span>to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :) 
</span><span>physical changes -- it can be melted, and oxidized 

</span><span>the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V</span>
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serg [7]
Urrrrrrrr nope now clue at all
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What is the difference between a group and a period
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3 years ago
An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa
sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

T_s = T_e \times \gamma

where c is the speed of light

v is the speed of the rock star moving

T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}

1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}

\sqrt{1-\dfrac{v^2}{c^2}} =0.556

squaring both side

1-\dfrac{v^2}{c^2}=0.3086

v^2=0.6914c^2

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v = 83.1 % of speed of light

7 0
3 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
3 years ago
Opposite electric charges will repel, and like charges will attract.
DochEvi [55]
As it stands now, that statement is false.

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#2).  Exchange the places of the words "repel" and "attract".

Either ONE of these changes will make the statement true.
Doing BOTH of these changes will leave it false.
5 0
3 years ago
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