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3 years ago

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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57.0- g tennis ball is traveling straight at a player at 21.0 m/ s. The player volleys the ball straight back at 25.0 m/ s. If t

Yakvenalex [24]

Answer:43.7 N

Explanation:

Given

Mass of tennis ball is =57 gm

velocity of ball=21 m/s

The player volleys the ball straight back at 25 m/s(i.e. in opposite of initial direction)

time of contact=0.06 s

We know change in momentum $(\Delta P) =Impulse(Fdt)$

initial momentum $=57\times 10^{-3}\times 21$

final momentum $=-57\times 10^{-3}\times 25$

change in momentum

$\Delta P=57\times 10^{-3}\times 21-(-57\times 10^{-3}\times 25)$

$\Delta P=57\times 10^{-3}\times 46=2.622 kg-m/s$

Now 2.622 $=F_{avg}\times 0.06$

$F_{avg}=43.7 N$

4 0

3 years ago

If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop a

MAVERICK [17]

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current $I_1$ = 100 μA

current $I_2$ = 1 mA

forward voltage $V_r$ = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.

Using the formula:

$I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

$I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

where;

$V_r$ = 0.7

$I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}$

$I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}$

$\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

$\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

Suppose n = 1

$V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV$

Then;

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}$

${e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}$

${\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}$

${\dfrac{V_r'}{nv_T}} =29.37$

$V_r'=29.37 \times nV_T$

$V_r'=29.37 \times 25.86$

$V_r'=759.5 \ mV$

$Vr' \simeq$ 760 mV

Thus, the voltage drop across this same diode will be 760 mV

3 0

3 years ago

according to newton's law of universal gravitation, in which of the following situations does the gravitational attraction betwe

Alex17521 [72]

Explanation:

The force acting between two masses is given by :

$F=G\dfrac{m_1m_2}{d^2}$ ........(1)

Where

G is the universal gravitational constant

$m_1\ and\ m_2$ are masses

d is the distance between two masses

It is clear from equation (1) that the gravitational attraction between the bodies always increases if the masses of bodies increases and when the separation between masses decreases.

So, the correct answer is "the masses increase, and the distance between the centers of mass decreases". This is because the force of gravitation is directly proportional to the masses and inversely proportional to the separation.

4 0

3 years ago

Describe how the boiling point of water on top of a mountain would be different from its boiling point at sea level.

Sonbull [250]

Answer:

At elevated altitudes, any cooking that involves boiling or steaming generally requires compensation for lower temperatures because the boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure.

Explanation:

7 0

3 years ago

The area of a piston of a force pump is 8 X 10⁴m². What force must be

prisoha [69]

Answer:

Force of $37.44 * 10^{8}$ Kgm/s^2 must be applied to the piston to raise oil to a height of 6.0 m