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ohaa [14]
2 years ago
14

An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byem

itting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?
Physics
1 answer:
ohaa [14]2 years ago
3 0

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

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How can we represent a narrow beam of light?
Yanka [14]

Answer:

It is easy to imagine representing a narrow beam of light by a collection of parallel arrows—a bundle of rays. As the beam of light moves from one medium to another, reflects off surfaces, disperses, or comes to a focus, the bundle of rays traces the beam's progress in a simple geometrical manner.

I hope it's helpful!

7 0
2 years ago
Read 2 more answers
A. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
Black_prince [1.1K]

Answer:

a

  n  = 23

b

  v  = 87377.95 \ m/s

Explanation:

From the question we are told that

   The diameter is d = 61\ nm =  61 *10^{-9} \ m

   

Generally the radius electron orbit  is mathematically represented as

      r = \frac{61 *10^{-9}}{2}

=>   r = 3.05*10^{-8} \  m

This radius can also be represented mathematically  as

      r =  n^2 *  a_o

Here n is the quantum number and a_o is  the Bohr radius with a value

    a_o =  0.0529 *10^{-9} \ m

So

   n  =  \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }

=>   n  = 23

Generally the angular momentum of the electron is mathematically represented as

          L  =  m * v *  r  =  \frac{n  *  h }{2 \pi}

Here  h is the Planck constant and the value is  h  =  6.626*10^{-34} J \cdot s

          m is the mass of the electron with values m  =  9.1*10^{-31} \  kg

         So

               v  =  \frac{23   *  6.626*10^{-34} }{2\pi * 9.1 *10^{-31}  * 3.05*10^{-8} }

                v  = 87377.95 \ m/s

       

3 0
3 years ago
Which characteristics describe the troposphere? Select two options.
lesantik [10]

Answer:

is the innermost layer,

has the highest air pressure

Explanation:

The troposphere is the lowest layer of Earth's atmosphere and site of all weather on Earth. The troposphere is bonded on the top by a layer of air called the tropopause, which separates the troposphere from the stratosphere, and on bottom by the surface of the Earth.

5 0
3 years ago
Read 2 more answers
La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p
liubo4ka [24]

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

4 0
2 years ago
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
2 years ago
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