It produces Heat Energy that can be used to drive steam turbines to generate Electricity.
The continental crust
Hope it helps!!
In stars more massive than the sun, the core temperature is hotter, which allows for fusion of more complex elements.
Most of the fusion occurs in the core.
In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.
Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.
Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.
After the core is almost entirely iron, the star is no longer in the Main Sequence.
So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.
Answer:
Explanation:
Let v be the velocity acquired by electron in electric field
V q = 1/2 m v²
V is potential difference applied on charge q , m is mass of charge , v is velocity acquired
2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²
v² = 844 x 10¹²
v = 29.05 x 10⁶ m /s
Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .
Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron
= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶
= 79.02 x 10⁻¹³ N .
Minimum force will be zero when electron moves along the direction of magnetic field .
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
