Answer:
The answer cannot be determined.
Explanation:
The energy of the diver when he hits the pool will be equal to its potential energy
, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.
The change in temperature
then will be

Where m is the mass of water in the pool, c is the specific heat capacity of water, and
is the added heat which in this case is the energy of the diver.
Since we do not know the mass of the water in the pool, we cannot make this calculation.
The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

Generally, the equation for Rate of flow of Liquid is mathematically given as

$$
Where dP is pressure difference r is the radius
is the viscosity of water
L is the length of the pipe


In $30s the quantity that flows out of the tube

In conclusion, the quantity that flows out of the tube

Read more about the flows rate
brainly.com/question/27880305
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Scientific theory because it's a theory it's already an answer but it might change depending on the condition.
b) between poles M1 and M2
Explanation:
From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.
The law of attraction between two magnetic poles states that:
<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s
Mathematically:
FM = K 
here r is the distance between the poles
FM is the magnetic force between the poles
M1 is the strength of the first magnetic pole
M2 is the strength of the second pole
K is the magnetic field constant
learn more:
magnetic pole brainly.com/question/2191993
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For every reaction, there is an equal and opposite reaction. This means that with every interaction, theres a pair of opposite forces on the object.