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3 years ago

How many carbon-14 atoms are left after 1 half-life

Physics

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

The time it takes for 14C to radioactively decay is described by its half-life. C has a half-life of 5,730 years. In other words, after 5,730 years, only half of the original amount of 14C remains in a sample of organic material. After an additional 5,730 years–or 11,460 years total–only a quarter of the 14C remains.

Explanation:

Hope this helps

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3. How is using a model to study cells helpful?

4vir4ik [10]

Answer:

Yes, it is very helpful.

Explanation:

It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.

I hope this helps!

6 0

3 years ago

How many protons, electrons, and neutrons are present in an atom of cr-24?

mote1985 [20]

The number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

An atom is the smallest unit of matter, consisting of the positively charged nucleus and the electrons which move around it. The atom can not be divided further.

The atom of a matter is made by three elements-

1) Neutron-Neutron is the element of an atom, which has zero charge.
2) Proton-Proton is the element of an atom, which has a positive charge.
3) Electron-Electron is the element of an atom, which has a negative charge.

The atom of cr-24 has the atomic number 24. The atomic number of atom is equal to the number of proton and electron in an atom. Thus,

$\rm Proton=24\\\rm Electron=24$

Now, if this atom has the mass number of 52 (most common). Then the number of neutron is,

$\rm Neutrons=Mass\; Number-Atomic\; Number\\\rm Neutrons=52-24\\\rm Neutrons=28$

Thus, the number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

Learn more about the atom here;

brainly.com/question/17545314

#SPJ4

8 0

2 years ago

Silver has a work function of 4.5 eV . Part A What is the longest wavelength of light that will release an electron from a silve

uranmaximum [27]

Answer:

λ = 2.7608 x 10⁻⁷ m = 276.08 nm

Explanation:

The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:

Work Function = hc/λ

where,

Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = longest wavelength capable of releasing electron.

Therefore,

7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)

7 0

3 years ago

A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort

Nutka1998 [239]

Answer:

(a) 1500 m

(b) 2827.43m

Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

$\Theta=\frac {\pi}{2}$

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, $l= \theta R$ , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

So, the distance covered by the child in 1 swing $= \frac {\pi}{2}\times 3 m=\frac{3\pi}{2}m$ .

In 1 cycle, there are 2 swings, so distance covered in 1 cycle $= 3 \pi m$ .

Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.

So, the total distance covered by the child

$= 3\pi\times 300 m= 2827.43 m$ .

3 0

3 years ago

The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e

Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ( $v$ ), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression: