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.
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
Answer:D
Explanation:
It was right o khan academy
Refer to the figure below.
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
Answer:
Earth attract the Moon with a force that is greater.
Explanation:
According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, F1 = Gm1m2/r²... 1
Let m1 be the mass of the earth and m2 be that of the moon
If the Earth is much more massive than is the Moon, the new force of attraction between them will become;
F2= G(2m1)m2/r²
F2 = 2Gm1m2/r² ... (2)
Dividing eqn 1 by 2 we have;
F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)
F1/F2 = Gm1m2/r²×r²/2Gm1m2
F1/F2 = 1/2
F2=2F1
This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)
