Answer: b
Explanation: did the quiz got it right
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
Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
The highest trophic level has the least available energy in kilojoules.
Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.
Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.
At the highest trophic level, the the least available energy in kilojoules in this food web is found.
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