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3 years ago

ANSWER ASAP PLEASE 15 POINTS

Physics

1 answer:

Sholpan [36]3 years ago

3 0

Answer:

SO4

Explanation:

I'm not sure if you're asking for more than one answer but S04 is one of them, you can search it up if you don't believe me

ion with the subscript of the atoms in a polyatomic ion. There is only ONE Cu and ONE S04, so get the charge for the Cu based on the S04. The formula is S04 , and there is only ONE S04 2, so Cu's charge here must be +2 for the compound to have an overall charge of zero. = copper (Il) ion S04 = sulfate ion then CuS04 = copper (Il) sulfate

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4 years ago

A. Una onda armónica que se propaga por un medio unidimensional tiene una frecuencia de 500Hz y una velocidad de propagación de

Natasha2012 [34]

Answer:

0.117 m

Explanation:

First of all, we can find the wavelength of the wave in the problem, by using the wave equation:

$v=f\lambda$

where:

v = 350 m/s is the speed of the wave

f = 500 Hz is the frequency of the wave

$\lambda$ is the wavelength

Solving for $\lambda$ ,

$\lambda=\frac{v}{f}=\frac{350}{500}=0.7 m$

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$\phi=2\pi$

is 0.7 m.

Here we want to find the distance between two points that have a difference of phase of

$\phi'=\frac{\pi}{3}$

So, we can set up the following rule of three:

$\frac{d}{\phi}=\frac{d'}{\phi'}$

where d' is the distance we are looking for. Solving for d',

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3 years ago

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}$

$d=\frac{1485*5.63}{2}\\d= 4180.3m$

wavelengt of sound is $\lambda$ = v/f

= (1485)/(5920)

= 0.251 m

7 0

3 years ago

The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

$R = \rho \frac{L}{A}$

Where,

R= Resistance of the conductor

$\rho =$ Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

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Using the equation from Resistance we have that at the muscle the value is:

$R_m = \rho \frac{L}{A}$

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At the same time we can make the same process to calculate the resistance of the fat, then

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$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

$R_f = \rho \frac{L}{A}$

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

$I= \frac{V}{R}$

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

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4 years ago