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3 years ago

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Literary Analysis: Sound Devices. The words jump and stunt are examples of which sound device?

1 answer:

Aleksandr-060686 [28]3 years ago

6 0

The answer is a. attliteration..................

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Describe sound as a wave.

EastWind [94]

Lol .. a what - a wave ? You just greeting somebody basically when your waving at someone I don’t really get the question but :) help it’s okay !

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3 years ago

If two opposing forces are equal, then the net force is 0 N.<br><br> true or false?

Fofino [41]

Answer:

false

Explanation:

6 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,

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

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3 years ago

When did the angels last win a world series

Colt1911 [192]

The last time those Angels won a world series was back in 2002. Go Nationals. ;)

8 0

3 years ago

Read 2 more answers

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th

NeX [460]

Answer:

3.14946 rad/s

Explanation:

$I_i$ = Intial moment of inertia

$I_f$ = Final moment of inertia

$\omega_i$ = Initial angular velocity

$\omega_f$ = Final angular velocity = $\dfrac{2}{1.33}\times 2\pi\ rad/s$

$\dfrac{I_f}{I_i}=\dfrac{1}{3}$

In this system the angular momentum is conserved

$L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s$

The angular velocity when the diver left the board is 3.14946 rad/s

3 0

3 years ago