Answer:
The combining of light nuclei is called nuclear fusion.
Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;
Here, C is the capacitance, 
is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;
Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.
Answer:
a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m
Explanation:
Mass of the dart = 0.02kg, the spring was compressed to 6cm
Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m
Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m
Work needed to compress the spring = 0.036J
b) the total energy stored in the spring = the work done to compress the spring = 0.036J
c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.
d) 1/2mv^2 = 0.036
mv^2 = 0.036*2
v^2 = 0.036*2 / 0.02 = 3.6
v = √3.6 = 1.897 approx 1.9m/s
e) kinetic energy of the dart = work done against gravity to get the body to height h
Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward
0.036 = 0.02 * 9.81 * h
0.036 / ( 0.02*9.81) = h
h = 0.18 m
Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:
B = μ₀I/(2πr)
B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line
Given values:
μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m
Plug in and solve for B:
B = 4π×10⁻⁷(170)/(2π(8.0))
B = 4.25×10⁻⁶T
The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:
4.25×10⁻⁶/(5.0×10⁻⁵)
= 0.085
= 8.5%
The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.
