Answer:
3.75 m/s
Explanation:
From the question given above, the following data were obtained:
Acceleration (a) = 1.5 m/s²
Initial velocity (u) = 0 m/s
Time (t) = 2.5 s
Final velocity (v) =?
a = (v – u) / t
1.5 = (v – 0) / 2.5
1.5 = v / 2.5
Cross multiply
v = 1.5 × 2.5
v = 3.75 m/s
Hence, the escape velocity of the squirrel is 3.75 m/s
Answer:
Because 'distance per second' is a velocity, not an acceleration.
Explanation:
Because 'distance per second' is a velocity, not an acceleration. For example, at 1 m/s an object is travelling a distance of 1 metre every second. But a rate of acceleration is a steady increase in velocity. So at 1 m/s^2, an object's velocity is increasing by 1 m/s every second.
Answer:
Explanation:
Unknown fork frequency is either
335 + 5.3 = 340.3 Hz
or
335 - 5.3 = 329.7 Hz
After we modify the known fork, the unknown fork frequency equation becomes either
(335 - x) + 8 = 340.3
(335 - x) = 332.3
x = 2.7 Hz
or
(335 - x) + 8 = 329.7
(335 - x) = 321.7
x = 13.3 Hz
IF the unknown fork frequency was 340.3 Hz,
THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz
IF the unknown fork frequency was 329.7 Hz,
THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz
