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aliina [53]
3 years ago
9

Indica 3 aplicações dos ultrassons. *

Physics
1 answer:
Ainat [17]3 years ago
3 0

Answer:

El ultrasonido se utiliza en muchos campos diferentes. Los dispositivos ultrasónicos se utilizan para detectar objetos y medir distancias. La ecografía o la ecografía se utilizan a menudo en medicina. En las pruebas no destructivas de productos y estructuras, el ultrasonido se utiliza para detectar defectos invisibles.

Espero que esto ayude, tenga un maravilloso día / noche, manténgase a salvo, felices fiestas y feliz navidad.

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What is one property of iodine
mamaluj [8]

Answer: I2

Explanation:

6 0
3 years ago
Part A:
svetlana [45]
Part A:
( 5,000 : 100 ) * 6 = 300 lit.
300 lit.  * 1.066 Euros/ lit. = 318.9 Euros
1 Euro = 1.20 USD
318.9 * 1.20 = 382.68 USD

Part B :
6 lit./100 km = 0.06 lit/1 km
1 gal = 3.7853 lit.
1 mile = 1.609344 km
0.06 : 3.7853 = 0.01585
... = 0.01585 gal / 1 km  / * 1.609344
= 0.0255 gal/mile
 
8 0
3 years ago
Read 2 more answers
PLEASE I NEED HELP ASAP!!!!!!!!!!
max2010maxim [7]

tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

3 0
3 years ago
Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s
Marrrta [24]
F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m  a₂ ⇒ a₁ = 2 a₂
 
since v = at + v₀  with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m

p₁ = p₂

8 0
3 years ago
Titanium is a metal used to make golf clubs. A rectangular bar of this metal measuring 1.96 cm x 2.19 cm x 2.63 cm was found to
masha68 [24]

4.012\frac{gr}{cm^3}

Explanation

the density of an object is given by:

\text{Density(d)}=\frac{mass(m)}{\text{volume(v)}}

Step 1

find the volume of the bar

a)find the volume of the rectangular bar.

the volume of a rectangular prism is given by:

\text{Volume}=\text{ length}\cdot widht\cdot depth

replace

\begin{gathered} \text{Volume}=(\text{ 2.63}\cdot2.19\cdot1.96)(cm^3) \\ \text{Volume}=11.289012(cm^3) \end{gathered}

Step 2

now,

Let

\begin{gathered} \text{Volume}=11.289012(cm^3) \\ \text{mass}=\text{ 45.3 gr} \end{gathered}

replace in the formula

\begin{gathered} \text{Density(d)}=\frac{mass(m)}{\text{volume(v)}} \\ d=\frac{45.3\text{ gr}}{11.289012(cm^3)} \\ d=4.012\frac{gr}{cm^3} \end{gathered}

therefore, the answer is

4.012\frac{gr}{cm^3}

I hope this helps you

4 0
1 year ago
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