Answer: Both Technician A and B
Explanation:
There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.
And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.
Total time in between the dropping of the stone and hearing of the echo = 8.9 s
Time taken by the sound to reach the person = 0.9 s
Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds
Initial speed (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time taken (t) = 8 seconds
Let the depth of the well be h.
Using the second equation of motion:

h = 313.6 m
Hence, the depth of the well is 313.6 m
Answer: Neither Sandra nor Marissa will be in her THR zone.
Explanation:
1) Actual pulse of both Sandra and Marissa : 144 bpm
2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm
3) Sandra's TRH is in the range 135 - 172 bpm.
Since 124 < 135, she will be below the range.
4) Marissa's TRH range is 143 - 176 bpm.
Since, 124 < 143, she is below the range
In conlusion, neither Sandra nor Marissa will be in her THR zone.
Answer:
The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7 m/s
Explanation:
The Doppler shift equation is given as follows;

Where:
f' = Required observed frequency = 20.0 kHz
f = Real frequency = 21.0 kHz
v = Sound wave velocity = 330 m/s
= Observer velocity = X m/s
= Source velocity = 0 m/s (Assuming the source is stationary)
Which gives;

330 -
= (20/21)*330
= 330 - (20/21)*330 = 15.7 m/s
The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7 m/s.
Answer:
6.8 m/s2
Explanation:
Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is
W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N
For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of
a = F/m = 13.6 / 2 = 6.8 m/s2
So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail