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Karo-lina-s [1.5K]
2 years ago
5

Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea

ch are measured. Both stones have the same volume, 0.15 cm3. However, stone A has a mass of 0.52 g, and stone B has a mass of 0.42 g. If diamond has a density of 3.5 g/cm3, could either of the stones be real diamonds?
Physics
1 answer:
dimaraw [331]2 years ago
6 0

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone =  0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond =  3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

density\density = \dfrac{mass}{volume}

density of stone A

\rho_A = \dfrac{0.52}{0.15}

\rho_A = 3.467\ g/cm^3

density of stone B.

\rho_B = \dfrac{0.42}{0.15}

\rho_B = 2.8\ g/cm^3

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

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Answer:

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

R = 2.9 ohms

And a battery of 1.5V is applied

V = 1.5 V

We want to find the rate of energy transformed

First we need to know what rate of energy is

Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

Then,

In electronic, the power dissipated by a resistor is given as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts

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No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.
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A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?​
ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut    <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

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It increases by 35% ......

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

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KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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