A cylinder with a movable piston contains 2.00 gg of helium, HeHe, at room temperature. More helium was added to the cylinder an
2 answers:
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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.
Answer:
a) Em₀ = 42.96 104 J
, b) = -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s