A softball is hit into the outfield during a game. The initial velocity is 33.52 m/s, and the ball is hit at an angle of 27°. Ho
1 answer:
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Answer:
The energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
So, the energy in its ground state is 10 meV. Hence, this is the required solution.
Answer:
(A) 7.9 m/s^{2}
(B) 19 m/s
(C) 91 m
Explanation:
initial velocity (U) = 0 mph = 0 m/s
final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s
initial time (ti) = 0 s
final time (t) = 4.8 s
(A) acceleration =
= = 7.9 m/s^{2}
(B) average velocity =
= = 19 m/s
(C) distance travelled (S) = ut +
= (0 x 4.8) + = 91 m