Waves transport energy along a medium without transporting matter. The amount of energy carried by a wave is related to the amplitude of the wave. Thus, the higher the wave is from the resting line, the more energy is put in and vice-versa.
Answer:
a)143.8 decays/minute
b)0.41 decays/minute
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of C-14= 5670 years
t= time taken to decay
Ao= activity of a living sample
A= activity of the sample under study
a)
0.693/5670 = 2.303/1000 log(162.5/A)
1.22×10^-4 = 2.303×10^-3 log(162.5/A)
1.22×10^-4/2.303×10^-3 = log(162.5/A)
0.53 × 10^-1 = log(162.5/A)
5.3 × 10^-2 = log(162.5/A)
162.5/A = Antilog (5.3 × 10^-2 )
A= 162.5/1.13
A= 143.8 decays/minute
b)
0.693/5670 = 2.303/50000 log(162.5/A)
1.22×10^-4 = 4.61×10^-5 log(162.5/A)
1.22×10^-4/4.61×10^-5 = log(162.5/A)
0.26 × 10^1 = log(162.5/A)
2.6= log(162.5/A)
162.5/A = Antilog (2.6 )
A= 162.5/398.1
A= 0.41 decays/minute
The impulse experienced by an object is the force. • time.
The momentum change of an object is the mass. • velocity change.
The impulse equals the momentum change.
Your impulse is 1.875 I hope this helps
Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
