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HACTEHA [7]
4 years ago
10

if a piece of sea floor has moved 50 km in 5 million years what is the yearly rate of sea-floor motion?

Physics
2 answers:
BartSMP [9]4 years ago
6 0
<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
Sonbull [250]4 years ago
5 0

50 km = 50,000 meters

           (50,000 meters) / (5 million years)

       =    (5 x 10⁴) / (5 x 10⁶)    (meters/year)

       =           1 x 10⁻²      meter/year

       =           1 centimeter per year              
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Answer:

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8 0
3 years ago
You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
lbvjy [14]

Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

=>    T_c  =  12 .1 ^oC

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3 years ago
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I = Q/t, as the charge increase , the current will also increase.

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