A )
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81 + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
Answer:
Explanation:
If the energy levels of n1 and n2 are high, the difference of their energy level will tend to zero thereby making the energy of the emitted photons very low.
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m
Vf = 480.83 m/s
a=v/t
t=v/a
t=480.83/34
t=14.142 s
Explanation:
It is given that,
Mass of lithium, 
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :


q is the charge on an electron

v = 78608.58 m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :



r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.