The half-life of Cobalt 60 is 5.27 years. This means that after this time, only 50 % of the initial amount of Cobalt is left. The relationship between the number of nuclei left at time t and the time, for a radioactive decay, is given by
where
is the number of nuclei left at time t,
is the number of nuclei at t=0, and
is the half-life. For cobalt,
.
We can re-arrange the formula as
and then
the problem asks for the time t at which only 20% of cobalt is left, that means the time t at which
therefore, using this into the previous equation, we get
so, after a time of 12.24 years, only 20% of cobalt is left.
Answer:
8) A hanging 7.5 kg object stretches a spring 1.1 m. What is the spring constant? 1 . 1 66. 82 NIMO 9) A spring has a constant of 875 N/m. What hanging mass will cause this spring to stretch 4.5 m? Iouliply then divide 875 X 4. 2- - (4 04. 7919) 10) A spring with a spring constant of 25 N/m is stretched 65 cm. What was the force used? 11) A 25 N force stretches a spring 280 cm. What was the spring constant? 12) A 75 N force stretches a spring 175 cm. What was the proportionality constant? 8) 66.82 N/m 9) 401.79 kg 10) 16.25 N 11) 8.93 N/m 12) 42.86 N/m
Explanation:
Answer:
The option that best describes the relationship between frequency and wavelength of electromagnetic waves is the third option
c. The frequency increases as the wavelength decreases
Explanation:
The frequency of an electromagnetic wave, 'f', is given by the following formula;
From the above formula, the frequency of an electromagnetic wave is inversely proportional to the wavelength, and therefore, a higher frequency, ('f', to increase), requires a lower wavelength, (λ, decreases)
The correct option is that the frequency of an electromagnetic increases as the wavelength of the wave decrease.
Answer:
1. Transform Boundary
2. Continental-continental convergent boundary
3. Oceanic-Oceanic Divergent boundary
Explanation:
Answer:
α = π/3
β = π/6
Explanation:
Use arc length equation to find the sum of the angles.
s = rθ
π/20 m = (0.1 m) (α + β)
π/2 = α + β
Draw a free body diagram for each sphere. Both spheres have three forces acting on them:
Weight force mg pulling down,
Normal force N pushing perpendicular to the surface,
and tension force T pulling tangential to the surface.
Sum of forces on A in the tangential direction:
∑F = ma
T − m₁g sin α = 0
T = m₁g sin α
Sum of forces on B in the tangential direction:
∑F = ma
T − m₂g sin β = 0
T = m₂g sin β
Substituting:
m₁g sin α = m₂g sin β
m₁ sin α = m₂ sin β
(1 kg) sin α = (√3 kg) sin (π/2 − α)
1 sin α = √3 cos α
tan α = √3
α = π/3
β = π/6
