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4 years ago

In a controlled experiment, which variable does the investigator change?

Physics

2 answers:

bagirrra123 [75]4 years ago

8 0

The investigator changes the independent variable

Bas_tet [7]4 years ago

7 0

Answer:

The manipulated variable is also known as the independent variable

Explanation:

When remembering this remember independent as in you independently change the outcome

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While practicing the trumpet you notice that every time you play a particular note a window in the room rattles. How can you exp

kherson [118]

The sound waves travel because of the pitch of the note so the window rattles.

7 0

3 years ago

How can air particles be trapped for investigation? (Subject: Science)

zhannawk [14.2K]

-- Find a clean jar that has a tight lid.
-- Take the lid off of the jar.
-- Wave the jar around for a while.
-- Put the lid back on the jar, tightly.

You now have a jar full of air and everything in the air.
You can take it into your laboratory and have your way with it.

3 0

3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor

arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

$PV = nRT$

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

$V_{1} = \frac{nRT_{1}}{P_{1}}$ (1)

La presión cuando T = 67 °C es:

$P_{2} = \frac{nRT_{2}}{V_{2}}$ (2)

Dado que V₁ = V₂ (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

$P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm$

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

4 0

3 years ago

A sphere is charged with electrons to

pishuonlain [190]

A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.

N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13

3 0

3 years ago