It appears to be a <span>spiral shape. </span>
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.
D = 1/f, where D is the power in diopters and f is the focal length in meters.
D=1/20
<u>D=0.05</u>
The velocity of the ball when it was caught is 12.52 m/s.
<em>"Your question is not complete it seems to be missing the following, information"</em>,
find the velocity of the ball when it was caught.
The given parameters;
maximum height above the ground reached by the ball, H = 38 m
height above the ground where the ball was caught, h = 30 m
The height traveled by the ball when it was caught is calculated as follows;
y = H - h
y = 38 - 30 = 8 m
The velocity of the ball when it was caught is calculated as;

Thus, the velocity of the ball when it was caught is 12.52 m/s.
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