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Y_Kistochka [10]
3 years ago
9

Diferencia entre transferencia y transformación de energía. Da ejemplos de cada caso.

Physics
1 answer:
Firdavs [7]3 years ago
6 0

Answer:

La transformación de energía es un proceso en el que la energía se intercambia entre un sistema y el medio ambiente en al menos dos formas de energía diferentes entre sí. Por ejemplo, un panel solar convierte la energía lumínica en energía eléctrica.

En cambio, en la transferencia de energía, esta no cambia su forma sino que es transmitida de un cuerpo a otro. El ejemplo más claro es el de la fogata, que transmite calor al medio ambiente a través de radiación.

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QUESTION: A study group of students is reviewing claims in an article that gamma rays are among the most dangerous type of elect
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B

Bias is an inclination towards one way of thinking due to inherent factors in our environment, such as how one was brought up.

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In science, biases, conscious or unconscious, tend to make experiments’ results to incline in particular way rather than being impartial. Peer- review of scientific journals by the scientific community is meant to be a quality assurance measure to detect such biases. The students should check reviews of the scientific community on the article and see if any biases have been pointed out.  Probably the institution that the scientist works for is in competition with another institution that deals with gamma rays.

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3 0
3 years ago
Water is made of two hydrogen atoms and one oxygen atom bonded together. Julia is describing how water undergoes a physical chan
FrozenT [24]

Everything that Julia said is correct EXCEPT the words
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6 0
3 years ago
The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is suf
kupik [55]

Answer:

(a) I_{1}=3.2*10^{-3}W/m^{2}

(b) \beta =95dB

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2}  }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2}  }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2}  }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta  =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2}  }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB

8 0
3 years ago
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