The answer is true. fluorescence demonstrates a sparkly like light beam and also can bounce off of other beams
In light of this, V=V 0 loge (r/r 0 ) Field E= dr dV =V 0(r0r) eE= r mV2 alternatively, reV0r0=rmV2. V=(m eV 0 r 0 ) \ s1 / 2mV=(m e V 0 r 0 ) 1/2 = constant mvr= 2 nh, also known as Bohr's quantum condition or Hermitian matrix.
Show that the eigenfunctions for the Hermitian matrix in review exercise 3a can be normalized and that they are orthogonal.
Demonstrate how the pair of degenerate eigenvalues for the Hermitian matrix in review exercise 3b can be made to have orthonormal eigenfunctions.
Under the given Hermitian matrix, "border conditions," solve the following second order linear differential equation: d2x/ dt2 + k2x(t) = 0 where x(t=0) = L and dx(t=0)/ dt = 0.
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Answer:
a. 5.23 m/s² b. 44.23 N
Explanation:
a. What is the centripetal acceleration of the hammer?
The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.
Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.
So a = rω²
= 1.4 m × (3.74 rad/s)²
= 5.23 m/s²
b. What is the tension in the chain?
The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.
So, T = 8.45 kg × 5.23 m/s²
= 44.23 N
