Answer: The major challenges are as
1) understanding of the plasma: Plasma is a soup like mixture of subatomic particles of different atoms nuclei and electrons that are shattered apart by the temperature at which plasma is formed. further research is needed to understand the behavior of plasma so that it can be put to a proper use.
2) Confinement of plasma: Once we get the plasma we need to hold it so that we can obtain heat from it to drive a steam turbine but the sheer temperature of plasma is in millions of Celsius thus currently making it impossible to confine conventionally. Scientists use a loop of electric and magnetic fields to keep it in circulatory like manner so that it can be studied.
3) finally to obtain electricity from the plasma it should be stable to produce electricity. But currently to obtain pressure, temperature so that we have a sustained supply is highly difficult in technical and economical aspects.
Inertial confinement: In order to get the nuclei of atoms close enough for fusion this type of method used compression of the nuclei into highly small volumes.This is accomplished by use of lasers which are directed towards the fuel pellets that implode and travel towards other nuclei making fusion possible. It's main advantage is that it requires lesser time to initiate fusion but the disadvantage being that a large power is used to fire the lasers and the lasers should all hit the small target.
Magnetic Confinement: In this method we use a magnetic and electric fields in a properly designed space to keep the plasma in motion. In motion the nuclei of the atoms come close enough to initiate fusion.It's advantage being less power is required to start the process as compared to inertial confinement and the disadvantage being that plasma confinement is currently not properly understood.
A homozygous dominant
B: homozygous recessive
C: heterozygous
I hope this helps you :)
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C = 
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C = 
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ = 
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Answer:
c) both containers slide off the turntable at the same turntable speed.
Explanation:
The greater the mass of the cylinder, the greater the centrifugal force acting on it. That makes up for the greater force needed to move the heavier cylinder.
