Answer: C Heat Transfer, E Addition of Volatiles and A Decompression
Explanation:
- Decompression melting takes place within Earth when a body of rock is held at approximately the same temperature but the pressure is reduced.
<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
I'd guess at valve B. more information about the interesting question would help.
To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

Where,
B = Magnetic Field
I = Current
L = Length
<em>Note:
is a direct adaptation of the vector relation
</em>
From Newton's second law we know that the relation of Strength and weight is determined as

Where,
m = Mass
g = Gravitational Acceleration
For there to be balance the two forces must be equal therefore

Our values are given as,
Diameter 
Radius 
Magnetic Field 
From the relationship of density another way of expressing mass would be

At the same time the volume ratio for a cylinder (the shape of the wire) would be

Replacing this two expression at our first equation we have that:



Re-arrange to find I


We have for definition that the Density of copper is
, gravity acceleration is
and the values of magnetic field (B) and the radius were previously given, then:


The current is too high to be transported which would make the case not feasible.