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3 years ago

5

You have a frustrating day and just feel like screaming; you grab a pillow to scream into. Which interaction of sound would take

place here (Refraction, reflection, and absorption) ,and how did the sound waves move?

1 answer:

klemol [59]3 years ago

3 0

Absorption Because the sound waves go into the pillow. Not refecting or going through

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what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three

Serjik [45]

Answer:

1kW = 1000W

600W = 0.6kW

Cost of electric bill = 0.6kWh × 24 × 30 × $0.05

= $21.60

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2 years ago

1) A rock is dropped from a cliff with a height of 200 m. With what velocity will the rock hit the ground

Komok [63]

A rock is dropped from a 200 m high cliff. How long does it take to fall (a) the first 100 m and (b) the last 50 m?

The basic equation you want is:

s=at22

Solving for t:

t=2sa−−−√

We’ll assume a=9.8 , so 2a−−√=14.9−−−√≈0.4518

So, for (a)s=100 , so t=0.4518100−−−√=4.518

The total time is 0.4518200−−−√≈6.389

The time to fall 150 m is 0.4518150−−−√≈5.533

So the time to fall the last 50 m is 6.389 - 5.533 = 0.856 seconds

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2 years ago

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Microwave ovens emit microwave energy with a wavelength of 12.1 cm. What is the energy of exactly one photon of this microwave r

EastWind [94]

Answer:

$1.64\cdot 10^{-24}J$

Explanation:

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 wavelenght of the photon

For the microwave photons in this problem,

$\lambda=12.1 cm=0.121 m$

so their energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.121 m}=1.64\cdot 10^{-24}J$

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3 years ago

How do charging and discharging compare? How can charging go unnoticed,

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2 years ago

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

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3 years ago

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