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3 years ago

How do you solve for X ? (2tan(x)-4)^3 = 27

Physics

1 answer:

bixtya [17]3 years ago

6 0

Ummmmmm in only in 5th grade this ain't for me, sorry

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A student read his right ear against his desk while Teacher talks loudly on the desk and can hear the tapping sound only through

sladkih [1.3K]

Answer:

The student hears the wave that is transmitted by the desk

Explanation:

Mechanical waves need a material medium to be able to be transmitted, in the case of sound waves, one of the most common media is air, but it is also transmitted in other media in this case, stationery is transmitted.

The student hears the wave that is transmitted by the desk

The speed of the wave is proportional to the density of the material, so the wave that the student hears arrives much faster through the desk than through the air

6 0

3 years ago

The _____________ variable is observed, measured, and affected by the independent variable.

Katen [24]

The dependent variable

6 0

3 years ago

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The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda_0$ = Threshold wavelength = 459 nm

Work function is given by

$W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV$

The work function W0 of this metal is 2.7067 eV

4 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

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4 years ago

35 miles North 6 feet down 136 MB 85 km SE

pantera1 [17]

The answer is 133 Se

4 0

3 years ago

Read 2 more answers