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2 years ago

6

A student sees a newspaper ad for an apartment that has 1730 square feet (ft2) of floor space. How many square meters of area ar

e there?

1 answer:

SVETLANKA909090 [29]2 years ago

6 0

The correct answer for this question is 160.89 m².

The square footage of a newspaper ad for an apartment is = 1730ft²

We know that one square foot equals 0.093 square meters.

We can determine the area in square meters using the above conversion.

Total square meters = 0.093×1730ft² (or) 1730÷10.7

= 160.89 m²

To know more about conversion calculations, click here:

brainly.com/question/6703327

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A) A wire made from iron with a cross-section of diameter 0.800 mm carries a current of 14.0 A. Calculate the "areal current den

Veseljchak [2.6K]

Answer:

A) ρ= $1.74x10^{26}$

B) μ= $1.68x10^{29}\frac{electron}{m^3}$

C) v= $1.03x10^{-3}\frac{m}{s}$

D)e= $8.99x10^-9$

Explanation:

A)

The magnetic field can be find knowing the current is the charge per second

β= $\frac{14eC*s}{1.6x10^{-19}C}\\$

β= 8.75x10^{19}e*s

Electron density

ρ= $\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}$

B)

μ= $\frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}$

μ $=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}$

C)

The drift speed using last information found

$v= \frac{J}{n*q} \\v= \frac{14A}{\pi*((0.40x10^-3)^2)*1.68x10^29*1.60x10^-19)} = 1.03x10^-3(\frac{m}{s} )$

D)

To compared the random thermal motion and the current's drift speed

$e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9$

8 0

3 years ago

PLEASE HELP ME!!! the equations are around the question itself.

mixas84 [53]

I’m so so sorry I wish I could help.... My best answer is no because they have different shapes but I am not sure

6 0

4 years ago

A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

$f_o = f_s\frac {(v+vs)}{v}$ and $f_o = f_s\frac {(v-vs)}{v}$ when moving towards observer

With $f_o$ of 76 then taking speed in air as 343 m/s we have

$76 = f_s\times\frac {(343-vs)}{343}$

$f_s=\frac {343\times 76}{343-v_s}$

Similarly, with $f_o$ of 65 we have

$65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}$

Now

$f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}$

v_s=27.76 m/s

Substituting the above into any of the first two equations then we obtain

$f_s=70.07 Hz$

4 0

3 years ago

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$

Explanation:

Given

$I_1=2 Amps$

$I_2=0.75 Amps$

Length of each wires $\left ( L\right )=5 m$

Distance between wires $\left ( r\right )=5 cm$

Force per unit length = $\frac{\mu_0I_1I_2}{2\pi r}$

$F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}$

$F'=6\times 10^{-6}$

Force for $L=5 m$

$F=3\times 10^{-5} N$

6 0

3 years ago

A 9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container?

andreev551 [17]

Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg

8 0

3 years ago