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Lubov Fominskaja [6]
3 years ago
11

What are carbon-12, carbon-13, and carbon-14?

Physics
2 answers:
Jet001 [13]3 years ago
6 0
The answer would be C. (:
malfutka [58]3 years ago
3 0

Answer: C. different isotopes of the same element

Explanation:

Isotopes are the atoms of same element having different number of neutrons. The number of protons is same in all the isotopes which means that all the atoms have same atomic number. Having different number of neutrons, these  atoms have different mass number but same atomic number.

Carbon-12, Carbon-13, and Carbon-14 have same atomic number - 6 but different mass number i.e have different number of neutrons.

Carbon-12 has 6 neutrons, carbon-13 has 7 neutrons and carbon 14 has 8 neutrons. These are different isotopes of the same element.

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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

                                            a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t  - 1/2 ( 9.81)t^2

  to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

      10 m/s * .639 s =<u> 6.39 m </u>

5 0
1 year ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
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What is needed to give a larger boulder a large acceleration?
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More force needs to be applied
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