Suppose two wagons, one with twice as much mass as the other, fly apart when a compressed spring that joins them is released. Th
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First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A