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lbvjy [14]
3 years ago
14

Between what depths does Earth's temperature increase the slowest?

Physics
1 answer:
bija089 [108]3 years ago
8 0
Temperature increases at a rate of approximately 20o C/km in the upper crust (first 10 km) but then the rate decreases to only approximately 0.3o C/km below 200km due to the homogenizing effect of mantle convection. A good approximation of the rate of pressure increase with depth is: 30 MPa/km in the crust and 35 MPa/km in the mantle.
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The primary coil of an ideal transformer has 100 turns and its secondary coil has 400 turns. if the ac voltage applied to the pr
Nana76 [90]

The formula used in calculations relating to transformers is:

<span>Secondary voltage (Vs)/ Primary voItage (VP) = Secondary turns (nS)/ Primary turns (nP)</span>

 

Substituting the given values to find for Vs,

Vs / 120 V = 400 turns / 100 turns

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6 0
4 years ago
-Para lograr que una pieza de 0,300 kg de cierto metal aumente su temperatura desde 40°C a 60 °C ha sido necesario suministrarle
marishachu [46]

Answer:

parte a) el calor específico es c = 0.383 J /(gr*K)

parte b) la temperatura inicial es T inicial= 52.72 °C

Explanation:

para el primer punto la formula para el calor Q es:

Q = m * c * ( T final - T inicial )

donde

m= masa de la pieza = 0.300 kg = 300 gr

Q = flujo de energía en forma de calor= 2299 J

c = calor específico

T final = temperatura final =40°C

T inicial = temperatura inicial = 60 °C

entonces

Q = m * c * ( T final - T inicial )

c = Q / [ m* ( T final - T inicial ) = 2299 J/[ 300 gr  * ( 60 °C - 40°C )]

= 0.383 J /(gr*K)

c = 0.383 J /(gr*K)

para el segundo punto usamos la misma formula

Q = m * c * ( T final - T inicial )

pero

m= 200 gr= 0.200 kg

c=459.8 J/(kg*K) , Q =20.900 J , T final = 280 °C

Q = m * c * ( T final - T inicial )

T inicial = T final - Q/(m*c)  =280 °C - 20.900 J/(459.8 J/(kg*K)* 0.200 kg) = 52.72 °C

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i hope this helped! :)
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Answer:

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Explanation:

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