Answer:
B. 6 s
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 30 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (30 m/s) t + ½ (-10 m/s²) t²
0 = 30t − 5t²
0 = 5t (6 − t)
t = 6
You would have 50 dollars
Answer:
the maximum frequency observed is 2.0044 10⁶ Hz
Explanation:
This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is
f ’=
the + sign is used when the observer approaches the source
typical speeds of a baby's heart stop are around 200 m / min
let's reduce to SI units
v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s
let's calculate
f ’= 2 10⁶ (
)
f ’= 2.0044 10⁶ Hz
f ’= 1,9956 10⁶ Hz
therefore the maximum frequency observed is 2.0044 10⁶ Hz
Answer:
The change in frequency hearded by an observer in the truck is f'= 636.36 Hz.
Explanation:
Vo= 36 m/s
Vf= 45 m/s
V= 342 m/s
f= 500 Hz
f'= f * ((V- (-Vo)) / (V - Vf)
f'= 636.36 Hz