Solution :
Mass of the particle = M
Speed of travel = v
Energy of one photon after the decay which moves in the positive x direction = 233 MeV
Energy of second photon after the decay which moves in the negative x direction = 21 MeV
Therefore, the total energy after the decay is = 233 + 21
= 254 MeV
So by the law of conservation of energy, we have :
Total energy before the decay = total energy after decay
So, the total relativistic energy of the particle before its decay = 254 MeV
Answer: The reactions are controlled to regulate energy output
Explanation:
Nuclear energy is produced when either two nuclei fuse to produce larger nucleus or a bigger nuclei splits into two smaller nucleus.
Uranium easily splits to produce nuclear energy. The reactions are controlled and energy output can be regulated. Nuclear energy in abundant yet controlled amount is available. Nuclear energy is mostly used in steam turbines to produce electricity.
The average current density in the wire is given by:

where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:

where r is the radius of the wire. In this problem,
, so the cross-sectional area is

and the average current density is

Answer:
Answer:
1.1 x 10^9 ohm metre
Explanation:
diameter = 1.5 mm
length, l = 5 cm
Potential difference, V = 9 V
current, i = 230 micro Ampere = 230 x 10^-6 A
radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m
Let the resistivity is ρ.
Area of crossection
A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2
Use Ohm's law to find the value of resistance
V = i x R
9 = 230 x 10^-6 x R
R = 39130.4 ohm
Use the formula for the resistance



ρ = 1.1 x 10^9 ohm metre
Explanation:
Answer:
v = 38.73 m/s
Explanation:
Given
Extension of the bow, x = 50 cm = 0.5 m
Force of the arrow, F = 150 N
Mass of the arrow, m = 50 g = 0.05 kg
speed of arrow, v = ? m/s
We start by finding the spring constant
Remember, F = kx, so
k = F/x
k = 150 / 0.5
k = 300 N/m
the potential energy if the bow when pulled back is
E = 1/2kx²
E = 1/2 * 300 * 0.5²
E = 0.5 * 300 * 0.25
E = 37.5 J
The speed of the arrow will now be found by using the law of conservation of energy
1/2kx² = 1/2mv²
kx² = mv²
v² = kx²/m, on substituting, we have
v² = (300 * 0.5²) / 0.05
v² = 75 / 0.05
v² = 1500
v = √1500
v = 38.73 m/s