amos duhh_leand.............................
Answer:
f = 25 lbs
m = 5 lbs
a =?
f = ma
25 = 5 a( divide both sides by 5)
a = 5(lbs)
Answer:
Part a)

Part b)

Part c)

Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have



So it will read same as that of the mass

Part b)
When elevator is decending with constant speed then we will have



So it will read same as that of the mass

Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have



Reading is given as



Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Answer:
Explanation:
Speed of the source of sound = v = 44.7 m/s
Speed of sound = V = 343 m/s
a) Apparent frequency as the train approaches = f = [V /(V -v) ] × f
= [343 / (343 - 44.7) ] × 415 = 477.18 Hz
Wave length = λ = v / f = 343 / 477.18 = 0.719 m
b) Frequency heard as the train leaves = f ' = [V / ( V + v) ] f
= [343 / { 343 + 44.7 ) ] x 415
= 367.2 Hz
Wavelength when leaving = v / f = 343 / 367.2 = 0.934 m