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musickatia [10]

3 years ago

7

A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. App

lying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm

1 answer:

monitta3 years ago

4 0

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

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A violin string that is 50.0 cm long has a fundamental frequency of 440 Hz. What is the

amid [387]

For a standing wave on a string, the wavelength is equal to twice the length of the string:
$\lambda=2 L$
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
$\lambda = 2 \cdot 0.50 m = 1.00 m$

And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
$v=\lambda f = (1.00 m)(440 Hz)=440 m/s$

5 0

3 years ago

Read 2 more answers

A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e $\dfrac{m}{2}$ .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

$mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v$

The velocity of the other fragment immediately following the explosion is v .

4 0

2 years ago

How does the sun's energy most directly influence precipitation in an area?

topjm [15]

The sun's energy influences climate in various ways. For example the latitudes at the equator receive more energy from the sun and therefore have warmer temperatures, On the other hand the sun's energy influences precipitation in a climate by driving the water cycle which determines precipitation.The sun is what makes the water cycle take place. That is the sun provides energy or heat to the earth; the heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds ( precipitation), that in turn give us rain

5 0

2 years ago

Select all that apply to electrons and energy levels. - Scientists have not yet determined exactly why electrons do not collapse

Elenna [48]

Answer:

- The limitation of the maximum number of electrons in a given energy level can be used to account for the periodic recurrence of properties as the number of electrons increases.

Explanation:

First - Scientists have not yet determined exactly why electrons do not collapse into the nucleus. FALSE: Scientists do know why electrons do not collapse. Since the beginins of quantum mechanics it's known that the energy at small scales is quantized, that means there only can be certain values meaning that the energy do not change continously. In the case of the electron, it can only have certain levels of energy, that means they do not radiate continously as the go arround the atom, instead it is only allowed to have a certain amount of energy in a given state therefore it can not lose energy continously collapsing into the nucleus.

Second - Electrons cannot be located between levels except when they are in the process of moving. FALSE: We can not say that a electron moves between energy levels, it only can exist in any of the levels, but never in between. Also, the electron in any of its possible energy lavels can not be located with complete certainty due to the uncertainty principle.

Fourth - Electrons have any random energy. FALSE: as exposed above the electrons can only have certain cuantized energy levels acordinly to the rules of quantum physics

Fifth - Electrons can be found between energy levels. FALSE: Like said before we can not say that a electron exists between energy levels, it only can exist in any of the allowed levels, but never in between.

Thirth (correct one) : - The limitation of the maximum number of electrons in a given energy level can be used to account for the periodic recurrence of properties as the number of electrons increases. TRUE: the maximum number of electrons allowed in a given energy level directly determines the tipe of bond an atom can made with another (this due to the number of electrons in the higest energy level), so for example the elements in the left of a given row of the periodic table tend to have ionic bonds, but in the other hand the elements on right side tend form more covalent bonds. And this characteristic directly correllate with diferent properties of the elements.

7 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago