velocity of the physics instructor with respect to bus

acceleration of the bus is given as

acceleration of instructor with respect to bus is given as

now the maximum distance that instructor will move with respect to bus is given as




so the position of the instructor with respect to door is exceed by

so it will be moved maximum by 3 m distance
Answer:
An interaction of one object with another object results in a force between the two objects. Thus, at-least two objects must interact for a force to come into play.
Answer:
0.073 N-m
Explanation:
i = 12 A, l = 0.8 m, B = 0.12 T
The circumference of the loop is 0.8 m.
Let r be the radius of the loop.
2 x 3.14 x r = 0.8
r = 0.127 m
Maximum Torque = i x A x B
Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m
Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω = 
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
Answer:
YESS well it is partly nessary but it depends on the situation
Explanation: