Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.
m₁u + m₂u = (m₁ + m₂) x 1/3 u
3m₁ + 3m₂ = m₁ + m₂
3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2
The ratio of their inertias is 2
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.
So we need to find the formula for magnetic field B using the current (I) and the distance from the probe (d). So, We know that the stronger the current I, the stronger the magnetic field B. That tells us that the I and B are proportional. Also we know that the strength of the magnetic field B is weaker as the distance d of the probe increases. That tells us that B and d are inversely proportional. So our formula should have B=(I/d)*c where c is a constant of proportionality. c=μ₀/2π where μ₀ is the permeability of free space. So finally our formula is B=(μ₀I)/(2πd).
Gravity is the correct answer.
Answer:
1.1x10^-2N
Explanation:
We have the change in momentum as
P = 0.3(4.5+12)g.mph
= 0.3x0.447x(4.5+12)x10^-3
Then the force that is exerted will be
F = p/∆t
∆t = 0.2
= 0.3x0.447x(4.5+12)x10^-3/0.2
= 0.1341x16.5x10^-3/0.2
= 1.1x10^-2
Therefore the force that was exerted is equal to 1.1x10^-2
