Answer:
At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop
Explanation:
The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s
To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g
we write out the equation of motion thus.
S = ut + 0.5at²
wgere
S = distance to come to complete stop
u = final velocoty = 0 m/s
a = acceleration = 60g = 60 × 9.81
t = time = 36 ms
as can be seen, the above equation calls up the given variable as a function of the required variable thus
S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m
At 60g, a person will travel a distance of 0.38 m before coing to a complete stop
Answer:
B) Power is the rate at which work is done
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters
Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s
Average velocity of Y = 546.41 / 30 = 18.2 m/s
Answer:
The separation of the 2 points should be 50.0 meters.
Explanation:
According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals
Applying the given values we get
thus the linear separation equals 
Applying the given values we get
A compound has a definite ratio of components
