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Sagot :
Answer:
0.67 ms
Explanation:
The total time it takes the sound to enter and leave the water is t and
t = t' + t" where t' = total time of sound in air = d'/v' where d' = total distance sound travels in water = 2D (since the sound waves passes through the water twice )where D = distance travelled from water to surface of copper block = 0.45 m and v' = speed of sound in water = 1500 m/s and t" = total time of sound in copper = d"/v" where d" = total distance sound travels in copper block = 2D' (since the sound waves passes through the copper block twice )where D' = thickness of copper block = 0.15 m and v" = speed of sound in water = 4600 m/s
So, t = t' + t"
t = d'/v' + d"/v"
t = 2D/v' + 2D'/v"
Substituting the values of the variables into the equation, we have
t = 2 × 0.45 m/1500 m/s + 2 × 0.15 m/4600 m/s
t = 0.90 m/1500 m/s + 0.30 m/4600 m/s
t = 0.0006 s + 0.000065 s
t = 0.000665 s
t = 0.665 ms
t ≅ 0.67 ms
Note that the sound wave passes through the water and copper block without refracting since it enters perpendicularly. Also, the sound wave passes twice through each of the copper block and water hence the total distance in each material is twice the initial distance or thickness.
Answer:
t = 6.63 10⁻⁴ s
Explanation:
The speed of sound is constant in a given material medium, so to calculate time we can use the uniform motion relationships
v = d / t
t = d / v
the speeds of sound in the three media are
air v = 343 m / s
water v₁ = 1493 m / s
copper v₂ = 5010 m / s
the distance traveled in the water on the round trip is
d = 2 x
d₁ = 2 0.45 0.90m
the time to travel this distance is
t₁ = d₁ / v₁
t₁ = 0.9 / 1493
t₁ = 6.028 10⁻⁴ s
the distance in the copper
d₂ = 2 0.15 m
d₂ = 0.30 m
the time to travel this distance in copper is
t₂ = d₂/ v₂
t₂ = 0.30 / 5010
t₂ = 5.988 10⁻⁵-5 s
the total travel time is
t = t₁ + t₂
t = 6.028 10⁻⁴ + 0.5988 10⁻⁴
t = 6.6268 10⁻⁴ s
t = 6.63 10⁻⁴ s
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