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8090 [49]
2 years ago
15

What conclusions can be drawn about the existence of carbon-12, carbon-13, and carbon-14?

Physics
2 answers:
SashulF [63]2 years ago
7 0
<span>These are isotopes of carbon and they all contain 6 protons and 6 electrons but each contains a difference number of neutrons - 6, 7, and 8 respectively.

^ This is the answer because an isotope changes the atomic mass, NOT atomic number. That means that the neutrons are changed, not the protons. </span>
mars1129 [50]2 years ago
3 0

Answer: These are isotopes of carbon and they all contain 6 protons and 6 electrons but each contains a difference number of neutrons - 6, 7, and 8 respectively.

Explanation: Isotopes of an element have similar number of protons but different number of neutrons.

General representation of an element is given as: _Z^A\textrm{X} where,

Z represents Atomic number  (for neutral atom) = number of protons= no of electrons

A represents Mass number  = number of protons + number of neutrons

X represents the symbol of an element

Thus  _6^{12}\textrm{C} ,   _6^{13}\textrm{C} and  _6^{14}\textrm{X} all contain same number of protons i.e 6 but contain (12-6) =6, (13-6)=7 and (14-6)=8 neutrons respectively.

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A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what
Lynna [10]

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

V_1 i_1 = V_2 i_2

here we know that

V_2 = 140 volts

i_2 = 3.5 A

i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1

V_1 = 49 volts

So primary coil voltage is 49 Volts

7 0
3 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
A mass of 1.00×10−6 μkg is the same as
tankabanditka [31]
I'm not sure but I know u is 10^6
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It is determined that a certain light wave has a wavelength of 3.012 x 10-12 m. The light travels at 2.99 x 108 m/s. What is the
gayaneshka [121]

Answer:

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Explanation:

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Answer:

Explanation:

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