By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre
1 answer:
Answer:
Mass of the pull is 77 kg
Explanation:
Here we have for
Since the rope moves along with pulley, we have
For the first block we have
T₁ - m₁g = -m₁a = -m₁g/4
T₁ = 3/4(m₁g) = 323.4 N
Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have
T₂ - m₂g = m₂a = m₂g/4
T₂ = 5/4(m₂g) = 134.75 N
T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)
∴
Mass of the pull = 77 kg.
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Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
(1)
<em>where : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>
To find A₀ we can use the following equation:
(2)
<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
<em>where : is the tree's carbon mass,
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Similarly, from equation (2) λ is:
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So, the initial activity A₀ is:
Finally, we can calculate the time from equation (1):
I hope it helps you!