Answer:
answer c. come easy dear.
Answer:
frequency of the first tuning fork can be 274 Hz or 266 Hz
Explanation:
Given data :
we have number of beats per second = 4
thus, it can also be written as
Beat frequency of the tuning fork, v = 4 Hz
Frequency of the tuning fork, f = 270 Hz
now, the beat frequency is actually the difference of the consecutive frequency.
thus, we have
Frequency of the first tuning fork as = 270 Hz + 4 Hz = 274 Hz
or
Frequency of the first tuning fork as = 270 Hz - 4 Hz = 266 Hz
Therefore, the frequency of the first tuning fork can be 274 Hz or 266 Hz
Answer:
-27.3 m/s
Explanation:
Given:
y₀ = 38 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2 (-9.8 m/s²) (0 m − 38 m)
v = -27.3 m/s
Or, you can solve with energy.
PE = KE
mgh = ½ mv²
v² = 2gh
v = -27.3 m/s