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2 years ago

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

2 answers:

7 0

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.

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.

You might be interested in

If 31.25

valentina_108 [34]

1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

I hope this helps.

4 0

2 years ago

Read 2 more answers

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

4 0

3 years ago

Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0

3 years ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

Five groups of four vectors are shown below. All magnitudes of individual vectors are equal. Please rank the groups based on the

valkas [14]

With the addition of vectors we can find that the correct answer is:

C) Q> P > R = S > T

The addition of vectors must be done taking into account that they have modulus and direction. The analytical method is one of the easiest methods, the method to do it is:

Set a Cartesian coordinate system

Decompose vectors into their components in a Cartesian system

Perform the algebraic sums on each axis

Find the resultant vector using the Pythagoras' Theorem to find the modulus and trigonometry to find the direction.

In this exercise indicate that the modulus of all vectors is the same, suppose that the value of the modulus is A.

We fix a Cartesian coordinate system with the horizontal x axis and the vertical y axis, we can see that we do not need to perform any decomposition, so we perform the algebraic sums

Diagram P

x-axis

x = 2A

y-axis

y = 2A

The modulus of the resulting vector can be found with the Pythagorean Theorem

P = $\sqrt{x^2+y^2}$

P = $\sqrt{4A^2 +4A^2 }= \sqrt{8} \ A$

P = 2 √2 A

Diagram Q

x-axis

x = 3A

y-axis

y = A

Resulting

Q = $\sqrt{x^2+y^2}$

Q = $\sqrt{9A^2 + A^2 }$

Q = $\sqrt{10} \ A$

Diagram R

x- axis

x = 0

y-axis

y = 2 A

Resulting

R = $\sqrt{4A^2 + 0}$

R = $\sqrt{4} \ A$

Diagram S

x-axis

x = 2 A

y-axis

y = 0

Resulting

S = 2A

Diagram T

x- axis

x = 0

y-axis

y = 0

Resultant T = 0

We order the diagram from highest to lowest

Q> P> R = S> T

When reviewing the different answers, the correct one is:

C. Q> P> R = S> T

Learn more about adding vectors here:

brainly.com/question/14748235

5 0

1 year ago