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snow_lady [41]
3 years ago
15

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

mes 10^{-5}~\text{N}\cdot\text{m}^2/\text{C}3×10 ​−5 ​​ N⋅m ​2 ​​ /C when the area is parallel to the sheet of charge. Find the charge density on the sheet.
Physics
1 answer:
deff fn [24]3 years ago
5 0

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

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The lowest note on a grand piano has a frequency of 28.1 Hz. It is a fixed string oscillating in its fundamental (longest wavele
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Answer:

2210.91 N

Explanation:

f = v/∧ = 1/2 √ T/ μ

where f= 28.1 Hz , T= tension ,

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mass density =  μ = 350÷1000/2.00

= 0.175kg/m

from  f = 1/2L √ T/ μ

make T the subject of the formula

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3 years ago
A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca
DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa

Calculate the first moment of area below point C

Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm

Now calculate shear stress at point C

=\frac{FQ}{It}

=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa

Calculate the principal stress at point C

\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa

Calculate the maximum shear stress at piont C

\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2}  \\\\=336.28KPa

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Answer:

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Explanation:

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Every time you eat a food, your body assimilates, transforms and takes advantage of its nutrients in order to live.

No single food can provide us with all the nutrients that the body needs. Only a healthy, varied and balanced diet will do it.

Types of nutrients :

Nutrients can be essential (the body cannot synthesize them on its own and we must acquire them through food) or non-essential (our body produces them from other components). According to their proportion they are divided into :

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Water is another essential nutrient, without which we could not live and that helps all the physiological processes of the body to take place.

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