All categories

3 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]3 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 a sound increases 5 dB, the sound becomes _______ times louder.

viva [34]

Well, I'll try to write the formula in a way that's not confusing,
but I'm afraid it might be slightly confusing anyway.

When you're working with dB, the basic rule is

   A change of 10 dB means either multiplying or dividing by 10 .

   Multiply something by 10 ==> it increases by 10 dB.
   Divide something by 10 ==> it decreases by 10 dB.

It turns out that another way to write all of this is . . .

   An increase of 10 dB ===> multiply the original amount by 10¹
   An increase of 20 dB ===> multiply the original amount by 10²
   An increase of, say, 7 dB ===> multiply the original amount by 10⁰·⁷

   A decrease of 10 dB ===> multiply the original amount by 10⁻¹
   A decrease of 30 dB ===> multiply the original amount by 10⁻³
   A decrease of, say, 13 dB ===> multiply the original amount by 10⁻¹·³

This question says: The sound increases by 5 dB .

That means the original 'intensity' or 'power' of the sound
is multiplied by
   10⁰·⁵ = √10 = about 3.162 (rounded) .

From the choices listed, the closest one is (c).

3 0

3 years ago

Read 2 more answers

HELP ME ILL MARK YOU THE BRAINLIEST

OleMash [197]

Option C: Later in the day, less power is developed in lifting each box

7 0

3 years ago

Read 2 more answers

Which of the following statements about matter is true? Mass and matter are always the same. Matter is made up of atoms and has

Zepler [3.9K]

The correct answer is Matter is made up of atoms and has mass.

7 0

3 years ago

PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

lesantik [10]

Responder:

20.3 ° C

Explicación:

Según la ley de Charles: cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa.

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$

sustituyendo tenemos

$\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\$

Cruz multiplicar tenemos

$T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3$

Temperatura delle braci 20.3°C

4 0

3 years ago

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

tatiyna

Answer: Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).

5 0

3 years ago