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Ira Lisetskai [31]
3 years ago
11

An ultrasound machine is being used to try to identify potential kidney stones. The machine is working properly and no kidney st

ones are found. Which statement best describes this situation? The generated sound wave will not be able to reflect off of the object. The sound wave will not be generated, so there will be no echoes received. The echoes will be made but will not be able to be received by the machine. The machine will not be able to process the received echoes into images.
Physics
2 answers:
Kazeer [188]3 years ago
7 0

Answer:

The generated sound wave will not be able to reflect off of the object.

Explanation:

bonufazy [111]3 years ago
4 0

Answer:

Explanation:

e d g e n u i t y 2020

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In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
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Explanation:

The gravitational force equation is the following:

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Your lungs aren’t the ones that make the sound
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Just the answer (PLSS)
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The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers.

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2 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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