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Setler79 [48]
3 years ago
14

We can know what elements make up a star by studying the star’s

Physics
1 answer:
mamaluj [8]3 years ago
7 0
Light
The light of the star is a direct result of the elements inside the star. Therefore, studying the light helps up understand the elements that make up a star.
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When fertilized what will the ovary grow into on a flower
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4 years ago
The intensity of light from a central source varies inversely as the square of the distance. If you lived on a planet only half
sertanlavr [38]

Answer:

the intensity will be 4 times that of the earth.

Explanation:

let us assume the following:

intensity of light on earth =J

distance of earth from sun = d

intensity of light on other planet = K

distance of other planet from sun = \frac{d}{2} (from the question, the planet is half as far from the sun as earth)

from the question the intensity is inversely proportional to the square of the distance, hence

  • intensity on earth : J = \frac{1}{d^{2} }

        Jd^{2} = 1 ... equation 1

  • intensity on other planet : K =  \frac{1}{(\frac{d}{2}) ^{2} }  (the planet is half as far from the sun as earth)

        K(\frac{d}{2}) ^{2} = 1 ....equation 2

  • equating both equation 1 and 2 we have

       Jd^{2} = K(\frac{d}{2}) ^{2}

       Jd^{2} = K\frac{d^{2}}{4}

       J = \frac{K}{4}

        K = 4J

       intensity of light on other planet (K) = 4 times intensity of light on earth (J)

5 0
3 years ago
Suppose you charges parallel plate capacitor with a dielectric between the plates using a battery and the. Remove the bater, iso
Makovka662 [10]

Answer:

A) increase.

Explanation:

  • By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:

        C = \frac{Q}{V} (1)

  • At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:

       C = ε₀*A / d  (2)

  • If the space between plates, is filled with a dielectric of dielectric constant κ, the above equation becomes:

       C =\frac{\epsilon_{0}*\kappa*A}{d} (3)

  • If the capacitor, once charged, is disconnected from the battery, charge must keep the same.
  • Now, if we remove the dielectric, as stated in (3) and (2), the capacitance C will decrease when removing the dielectric.
  • From (1) if C decreases, and Q remains constant, in order to keep both sides of the equation equal each other, V (the potential difference between plates), must increase.    
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By how much would 1500j of heat energy raise the temperature of 0.50kg of aluminum​
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pls what is the specific heat capacity of water

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