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Evgen [1.6K]
3 years ago
13

Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at

0 degrees C. Which option best describes this information?
These are the physical properties of water.
These are the chemical properties of water.
These are the physical changes water undergoes.
These are the chemical changes water undergoes.
These are the molecular changes water undergoes.
Physics
1 answer:
BARSIC [14]3 years ago
4 0

Answer: Option (c) is the correct answer.

Explanation:

Physical properties are the properties in which there is no change in chemical composition of a substance. On the other hand, chemical properties are the properties which change the chemical composition of a substance.

For example, when water boils at 100 ^{o}C then it changes into vapor state whereas when water freezes at 0^{0}C then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

Hence, we can conclude that best option describing given information is that these are the physical changes water undergoes.

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What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
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Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

Wavelength = \frac {h}{mv}

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}

Wavelength = \frac {6.6262 * 10^{-34}}{4.859}

Wavelength = 1.36 * 10^{-34} meters

7 0
2 years ago
an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi
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for this we apply, Heisenberg's uncertainty principle.

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the formula is,

Δp * Δx = h/4π

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by rearranging,

Δx = h / 4π * m(e).Δv

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2

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for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 /2.05

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therefore, we can say that the lower limits are   0.011 X 10⁻³  m for the electron and  3.23 X 10⁻³² m   for the bullet

To know more about bullet problem,

brainly.com/question/21150302

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
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Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

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now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

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now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

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by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

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8 0
3 years ago
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ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

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The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

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\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

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3 years ago
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