A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
Answer:
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Claudius Ptolemy. Bartolomeu Velho, Public Domain. ...
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Johannes Kepler. NASA Goddard Space Flight Center Sun-Earth Day. ...
Galileo Galilei. NASA
Answer:
W = F * s
Work done equals applied force * distance traveled
Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled
μ = coefficient of kinetic friction
F cos θ = force applied to motion of sled
s = distance traveled
[μ M g (1 - sin θ)] cos θ * s = work done in moving sled
Note that F = μ M g if applied force is in the horizontal direction
Answer:
The fraction of its energy that it radiates every second is
.
Explanation:
Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

Given that,
Kinetic energy = 6.2 MeV
Radius = 0.500 m
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula

Put the value into the formula


We need to calculate the rate at which it emits energy because of its acceleration is

Put the value into the formula


The energy in ev/s


We need to calculate the fraction of its energy that it radiates every second


Hence, The fraction of its energy that it radiates every second is
.