The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions
Answer:
A. False
B True
C. False
D.False
E. True
F. False
G. False
H. False
I. True
Explanation:
A. False: The system being analyzed consists of the bug and the car. These are the two bodies involved in the collision.
B. True: The system being analyzed consists of the bug and the car
C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car
D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.
E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.
F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision
G. False: The car barely changes in velocity since the mass of the bug is small.
H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.
I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity
Answer:
The energetic order third excited > seconds excited > first excited > ground
Explanation:
Bohr's atomic model includes an integer that is responsible for the quantization of energy
E = - 13.6 /n² eV
Where n is the so-called principal quantum number, it is an integer with values of
n = 1, 2, 3… inf
For n = 1 have the so-called ground state that is the one with the lowest energy, for the minus sign in the formula E1 = -13.6 eV
For n = 2 it is the first excited state has the most energy than the ground state,
E2 = 13.6 / 4 = -3.4eV
For n = 3 we have the second excited state, E3 = -13.6 / 9 = -1.5 eV
For n = 4 we have the third excited state, E4 = -0.85 eV
The energetic order of the states is
E4 > E3 > E2 >E1
