HEY CAN ANYONE PLS ANSWER DIS!!!!!!!!!!!!

Suppose wave pulses in an aquarium are produced by a mechanical motor that moves a bob up and down at the surface. The setup use

-Dominant- [34]

Answer:

kinetic energy in each pulse = 9.5J

Explanation:

The concept used here is that of work, energy and power.

Power P = Work /Time

Work = energy in this case

E = Power x time = Pt

but Power = 10W and t = 1.9s

Energy E = 10W x 1.9s = 19J

Conventionally, overall energy = kinetic + potential

Hence kinetic energy in each pulse = half of the total energy = 0.5 x 19

energy = 9.5J

3 0

3 years ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

lys-0071 [83]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)Does not affect the long term.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$

This is linear equation so integration factor ,I

$I=e^{\int kdt}$

$I=e^{kt}$

Now by using linear equation property

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)

at t= 0

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )$

So the initial condition does not affect the long term.

5 0

3 years ago

A bag is dropped onto the street from the roof of a 32 m high building. How long does it take for the bag to fall?

gizmo_the_mogwai [7]

Answer:

Explanation:

h = ½at²

t = √(2h/a)

t = √(2(32)/9.8)

t = 2.5555062... 2.6 s

4 0

3 years ago

Estimate the peak wavelength for radiation from ice at 273 k.

Andrews [41]

<h2>Answer: 10615 nm</h2>

Explanation:

This problem can be solved by the Wien's displacement law, which relates the wavelength $\lambda_{p}$ where the intensity of the radiation is maximum (also called peak wavelength) with the temperature $T$ of the black body.

In other words:

<em>There is an inverse relationship between the wavelength at which the emission peak of a blackbody occurs and its temperature.</em>

Being this expresed as:

$\lambda_{p}.T=C$ (1)

Where:

$T$ is in Kelvin (K)

$\lambda_{p}$ is the <u>wavelength of the emission peak</u> in meters (m).

$C$ is the <u>Wien constant</u>, whose value is $2.898(10)^{-3}m.K$

From this we can deduce that the higher the black body temperature, the shorter the maximum wavelength of emission will be.

Now, let's apply equation (1), finding $\lambda_{p}$ :

$\lambda_{p}=\frac{C}{T}$ (2)

$\lambda_{p}=\frac{2.898(10)^{-3}m.K}{273K}$

Finally:

$\lambda_{p}=10615(10)^{-9}m=10615nm$ This is the peak wavelength for radiation from ice at 273 K, and corresponds to the<u> infrared.</u>

8 0

3 years ago

R: Law of Conservation of Energy Pre-Write

Bad White [126]

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

KE=0.5*m*v^2

v=√[KE/(0.5*m)]=3.06 m/s

3 0

3 years ago