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-Dominant- [34]
3 years ago
5

What is the age of a sample if there are 6 grams of an element that have not decayed outof an original 24 grams and has a half-l

ife of 8yrs a. 8yrs b. 50yrs c. 29yrs d. 16yrs
Physics
1 answer:
amid [387]3 years ago
8 0

Answer:

d ) 16 years

Explanation:

If n be number of half years passed then

[tex]\frac{m}{m_0} =\sqrt[n]{\frac{1}{2} }[/tex]

m is mass undecayed m₀ is original mass

Here m = 6 g

m₀ =24 g

Putting these values

\frac{6}{24} =\sqrt[n]{\frac{1}{2} }

n = 2 half lives

age = 2 x 8 = 16 years

= 16 years

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On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
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Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

5 0
3 years ago
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