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:
1027.2 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32.2 ft/s


The height the tomato would fall is 450+577.2 = 1027.2 m
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by

where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi
num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched
Answer:
126.99115 g
Explanation:
50 g at 90 cm
Stick balances at 61.3 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)

The mass of the meter stick is 126.99115 g