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:
B
Explanation:
if you sit up straight you will have a proper posture
Answer:
His average speed is 45000
Answer:
q = -1.61x10⁻¹⁷ C
Explanation:
The charge of the particle can be found using the definition of the work done by electric force:
(1)
<u>Where</u>:
q: is the charge
ΔV: is the difference in electric potential
The work is also equal to:
(2)
<u>Where</u>:
and
are the electric potential energy of the points A and B, respectively.
Now, by conservation of energy we have:
(3)
<u>Where</u>:
and
are the kinetic energy of the points A and B, respectively.
Rearranging equation (3):


Solving the above equation for q:
Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.
I hope it helps you!