Answer:
Apply high pressure and high temperature to the rock
Explanation:
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Answer:
<em>The flux through the sphere will remain the same, and the magnitude of the electric field will increase by four times.</em>
Explanation:
The electric flux is the number of electric field, passing through a given area. It is proportional to the electric field strength and the area through which this field passes.
If the radius of the sphere is halved, the area of the sphere will reduce by square of the reduction, which will be four times. The electric field lines will become closer together, or technically increase by a fourth of its initial value. The resultant effect is that the electric flux will remain the same.
If originally,
Φ = EA cos∅
where Φ is the electric flux through the sphere
E is the electric field on the sphere
A is the area of the sphere.
If the area of the sphere is reduced to half, then,
the area reduces to A/4,
and the electric field increases to be 4E on the sphere.
The flux now becomes
Φ = 4E x A/4 cos∅
which reduces to
Φ = EA cos∅
which is the initial electric flux on the sphere.
Answer:
False
Explanation:
Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).
Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).
Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.
The diameter of the wire is 2.8 * 10^-3 m.
<h3>What is the length?</h3>
Mass of the wire = 1.0 g or 1 * 10^-3 Kg
Resistance = 0.5 ohm
Resistivity of copper = 1.7 * 10^-8 ohm meter
Density of copper = 8.92 * 10^3 Kg/m^3
V = m/d
But v = Al
Al = m/d
A = m/ld
Resistance = ρl/A
= ρl/m/ld =
l^2 = Rm/ρd
l = √ Rm/ρd
l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3
l = 1.82 m
A = πr^2
Also;
A = m/ld
A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3
A = 6.2 * 10^-5 m^2
r^2 = A/ π
r = √A/ π
r = √6.2 * 10^-5 m^2/3.142
r = 1.4 * 10^-3 m
Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m
Learn more about resistivity:brainly.com/question/14547003
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Missing parts;
Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?
