Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
It is a field of study that make direct use of phenomena that is "quantum-mechanincal", such as superposition and entanglement. It's used to perform operations on data
Convert 220 lb to kg.
220/2.2 = 100kg.
W = Fd (In this case, F is the weight)
W = (100)(2)
W = 200J
P = W/t
P = (200)/(1.2)
P = 166.67W
More info? I think the question is incomplete. Although, I believe the first 2 blanks are "rises"
