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kondaur [170]
4 years ago
13

A characteristic of matter that can be observed without changing the identity of the matter is a_______________.

Physics
1 answer:
ludmilkaskok [199]4 years ago
3 0

Answer: A characteristic of matter that can be observed without changing the identity of the matter is a physical property.

Explanation:

A property that does not bring any change in chemical composition of a substance are known as physical properties.

For example, shape, size, mass, volume, density, etc of a substance are all physical properties.

Whereas a property that changes chemical composition of a substance is known as chemical property.

For example, precipitation, reactivity, toxicity etc are chemical property.

A chemical change cannot be reverted back.

Therefore, we can conclude that a characteristic of matter that can be observed without changing the identity of the matter is a physical property.

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How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy
Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

u \:  =  \:  \frac{kq1q2}{d}

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0
4 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
What is the mass of an object that has a weight of 2867 N?
Gnom [1K]

Answer:

F = M a

W = M g     equivalent equation to express weight of object of mass M

M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg

7 0
3 years ago
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth
Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2

here we know that

\omega^2 = \frac{k}{m_1 + m_2}

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

\omega'^2 = \frac{k}{m_1}

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0
3 years ago
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