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3 years ago

How wood helps in home insulation plz answer fastplz answwweeeeeeerrrrrrr

2 answers:

8 0

Wood is a natural insulator due to air pockets within its cellular structure, which means that it is 15 times better than masonry, 400 times better than steel, and 1,770 times better than aluminum. In addition, lightweight wood framing methods allow easy installation of additional fibre or foil insulation. i hope this is better

Usimov [2.4K]3 years ago

3 0

Wood is a poor conductor and therefore a good insulator keeping heat inside

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A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power.

tatyana61 [14]

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

6 0

3 years ago

The hockey stick hits the puck forward. Give the reaction force

weqwewe [10]

I think the puck pushes the stick backwards

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3 years ago

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jarptica [38.1K]

Speed = distance / time
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7 0

3 years ago

Which material is used to reduce friction ?

Iteru [2.4K]

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Make the surfaces smoother. ...

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6 0

3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago