Answer:
reduce the velocity of collision
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
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Answer:
(2x + 4)(x - 4)=2x^2-4x-16
Answer:
Height of cliff = S = 20 m (Approx)
Explanation:
Given:
Initial velocity = 8 m/s
Distance s = 16 m
Starting acceleration (a) = 0
Computation:
s = ut + 1/2a(t)²
16 = 8t
t = 2 sec
Height of cliff = S
Gravitational acceleration = 10 m/s
S = 1/2a(t)²
S = 1/2(10)(2)²
Height of cliff = S = 20 m (Approx)
