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Nuetrik [128]
3 years ago
6

Atoms of element A decay to atoms of element B with a half-life of 20,000 years. If there are 10,000 atoms of A to begin with (a

nd 0 atoms of B), how long will it take for there to be 2,500 atoms of A? A) 20,000 years B) 40,000 years C) 60,000 years D) 100,000 years
Physics
2 answers:
Artyom0805 [142]3 years ago
7 0

Answer:

B) 40,000 years

Explanation:

You would have only 2500 of A left after 40,000 years. This represents 2 half-lives. After 20,000, ther would be 5,000 of A left and after another 20,000, it would be reduced to 2,500.

oee [108]3 years ago
4 0

T = half life period of decay for atoms of element A = 20,000 years

N₀ = initial number of atoms of element A = 10,000 atoms

N = final number of atoms after time "t" = 2500 atoms

t = time of decay = ?

λ = decay constant = ?

decay constant is given as

λ = 0.693/T

λ = 0.693/20,000

λ = 0.00003465 years⁻¹

atoms after decay for time "t" is given as

N = N₀ e^{-\lambda t}

inserting the values

2500 = (10000) e^{-(0.00003465) t}

t = 40,000 years

so correct choice is

B) 40,000 years


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F_1=2k\times d

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3 years ago
A person hits a 45-g golf ball. The ball comes down on a tree root and bounces
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Answer:

Maximum height, h = 10 m          

Explanation:

It is given that,

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We need to find the height the ball will rise  after the bounce. It is based on the conservation of energy such that,

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h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m

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5 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

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focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

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2 Bikers Rode at a constant speed on a 150 meter track.The data here show each bikers distance for a certain part of the race.Wh
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3 years ago
Read 2 more answers
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b
jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

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ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

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 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0
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