Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
Hence, this is the required solution.
convergent and counterclockwise
hope it helps :)
Answer:
V= A ω maximum KE of object in SHM
V2 / V1 = .958 ratio of amplitudes since ω is constant
KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2
KE2 / KE1 = .958^2 = .918
So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle
