Answer: 2.72 metres
Explanation:
Given that:
frequency of sound F = 123 Hz. wavelength of sound in the air = ?
speed of sound in air V = 334 m/s
Recall that wavelength is the distance covered by the wave after one complete cycle. It is measured in metres, and represented by the symbol λ.
So, apply V = F λ
λ = V /F
λ = 334m/s / 123Hz
λ = 2.72m
Thus, the wavelength of this sound in the air is 2.72 metres
Fnety = (FT)(sin 32°) – Fg
Or the answer B, I checked it.
Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
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Momentum question. This is an inelastic collision, so
m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s
0.67m/s