<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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Answer:
3.03 m/s²
Explanation:
m₂ > m₁, so m₁ will move up, m₂ will move down, and m₃ will move to the right.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₂ − m₂g = m₂(-a)
T₂ = m₂g − m₂a
Sum of the forces on m₃ in the y direction:
∑F = ma
N − m₃g = 0
N = m₃g
Sum of the forces on m₃ in the x direction:
∑F = ma
T₂ − T₁ − F = m₃a
T₂ − T₁ − Nμ = m₃a
Substituting and solving for acceleration:
(m₂g − m₂a) − (m₁g + m₁a) − (m₃g)μ = m₃a
m₂g − m₂a − m₁g − m₁a − m₃gμ = m₃a
g (m₂ − m₁ − m₃μ) = (m₁ + m₂ + m₃) a
a = g (m₂ − m₁ − m₃μ) / (m₁ + m₂ + m₃)
Given m₁ = 4 kg, m₂ = 12 kg, m₃ = 6 kg, and μ = 0.2:
a = 9.81 (12 − 4 − 6×0.2) / (4 + 12 + 6)
a = 3.03 m/s²
Answer:
its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.
Explanation:
When a scientist has a curiosity about something he carried out a research. and when their is a demand of something in society that time scientific research is carried out. Therefore its true that a scientific endeavor is driven by simple curiosity or societal demand.
For example
in society, there is demand of a medicine which can completely kill the cancer and a scientist has curiosity to know how to kill cancer cell. In this way a scientific endeavor for cancer medicine can be carried out by both simple curiosity as well as societal demands.
Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
horizontal component=fcostita
=150cos60
use calculator to evaluate it
for vertical=fsintita
=150sin60