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3 years ago

How to calculate sound of an echo

Physics

1 answer:

bonufazy [111]3 years ago

5 0

by an echo meter

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#Genius kudi

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Yosemite falls, in california, has a total height of 73,900 cm. what is this height in meters?

Hunter-Best [27]

Yosemite falls has a total height of 73,900 cm it means it heights 739 meters

6 0

3 years ago

If the torque on an object adds up to zero Group of answer choices the object is at rest. the object cannot be turning. the obje

garik1379 [7]

Answer:

the body has linear acceleration, but cannot rotate

Explanation:

Let's analyze the system

If the torque is zero, the two forces are the same magnitude, but applied to each side of the body in such a way that the torque cancels the punch of the other. Therefore the body cannot turn

The two forces go in the same direction so the object can have linear acceleration

The object is at rest because it has a force in the same direction, but in the opposite direction.

therefore the correct answer is:

the body has linear acceleration, but cannot rotate

8 0

3 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

6 0

3 years ago

What is another term for a pull on an object? O A. Acceleration O B. Speed O c. Force O D. Velocity

mihalych1998 [28]

Answer:

Acceleration

Explanation:

3 0

3 years ago

Read 2 more answers

An object is placed at a certain distance on the principal axis of a concave mirror. The image is formed at a distance of 30cm f

Nikolay [14]

the object distance is 10cm

6 0

3 years ago

How to calculate sound of an echo ​

How to calculate sound of an echo