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3 years ago

Yaşam alanlarında enerji tasarrufu

Physics

1 answer:

siniylev [52]3 years ago

4 0

(Forgive me if this is wrong I don't speak Turkish)

Enerji tasarrufu nasıl yapılır

Beklemedeki cihazları kapatın.

Akıllı bir termostat takın.

Termostatınızı kısın.

Verimli cihazlar satın alın.

Yeni bir kazan takın.

Giysileri daha düşük sıcaklıkta yıkayın.

Su konusunda daha akıllı olun.

Çift cama yatırım yapın.

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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

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Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

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3 years ago

If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a

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Answer:

K'=4K

Explanation:

The electric potential energy is given by :

$E=\dfrac{1}{2}kx^2$

Where

k is spring constant

x is compression or extension in the spring

If the displacement of a horizontal mass-spring system is doubled, x'= 2x

New elastic potential energy :

$E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K$

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3 years ago

What is 18 degrees celsius in fahrenheit ?

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18 degree is equal to 64.4 fahrenheit

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3 years ago

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

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Answer:

Explanation:

Given

mass of block $m=5.7\ kg$

at $t=0 s$

displacement is $x=-0.7\ m$

velocity $v=-0.8\ m/s$

acceleration $a=2.7\ m/s^2$

suppose $x=A\cos (\omega t+\phi )$ is the general equation of SHM

where A=amplitude

$\omega$ =natural frequency of oscillation

therefore velocity and acceleration is given by

$v=-A\omega \sin (\omega t+\phi )$

$a=A\omega ^2\cos (\omega t+\phi )$

for t=0

$-0.7=A\cos (\phi )---1$

$v=-0.8=-A\omega \sin(\phi)---2$

$a=2.7=-A\omega ^2\cos(\phi )----3$

divide 1 and 3 we get

$\omega ^2=\frac{27}{7}$

$\omega =\sqrt{\frac{27}{7}}$

Now square and 1 and 2 we get

$(0.7)^2+(\frac{0.8}{\omega })^2=A^2$

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$A=0.81\ m$

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3 years ago

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Yaşam alanlarında enerji tasarrufu​

Yaşam alanlarında enerji tasarrufu