This is the part of where you can easily slip away
Answer:
Explanation:
F = kQq/r²
r = √(kQq/F)
a) r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m
r = 40 mm
b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m
r = 42 mm
Answer:
Explanation:
a) Linear density is greater on the left side because the velocity of wave in a string is inversely proportional to the linear mass density of the string
b) We should start pulse from the left hand side so the reflected wave does not get inverted because the wave traveling from the denser to lighter medium gets reflected in the same phase.
Answer:
Approximately 
, assuming friction between the vehicle and the ground is negligible.
Explanation:
Let 
denote the mass of the vehicle. Let 
denote the initial velocity of the vehicle. Let 
denote the spring constant (needs to be found.) Let 
denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy (
) of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy (
) stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial 
of the vehicle should be equal to the 
of the vehicle. In other words:
.
Rearrange this equation to find an expression for , the spring constant:
.
Substitute in the given values 
, 
, and 
:
