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3 years ago

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What is the action force responsible for sea floor spread?

2 answers:

Alexxx [7]3 years ago

8 0

Answer: The motivating force for seafloor spreading ridges is tectonic plate slab pull at subduction zones, rather than magma pressure, although there is typically significant magma activity at spreading ridges. ... At a spreading center, basaltic magma rises up the fractures and cools on the ocean floor to form new seabed.

Explanation: Hope This Is Helpful

ollegr [7]3 years ago

4 0

Answer:

Seafloor spreading helps explain continental drift in the theory of plate tectonics. When oceanic plates diverge, tensional stress causes fractures to occur in the lithosphere. The motivating force for seafloor spreading ridges is tectonic plate slab pull at subduction zones, rather than magma pressure

Explanation:

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when a person strikes a match, potential energy in the match is transformed into which types of energy?

natulia [17]

Answer:

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mafiozo [28]

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3 years ago

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

nirvana33 [79]

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

$q=mC(T_2-T_1)$

where,

q = heat released = $1.0\times 10^3J=1000J$

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

$T_2$ = final temperature = ?

$T_1$ = initial temperature = 20°C

Putting values in above equation, we get:

$1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC$

Hence, the final temperature of the coffee is 43.9°C

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3 years ago

When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu

olga2289 [7]

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC = C = 0,50 J/gºC

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