You must observe the object twice.
-- Look at it the first time, and make a mark where it is.
-- After some time has passed, look at the object again, and
make another mark at the place where it is.
-- At your convenience, take out your ruler, and measure the
distance between the two marks.
What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.
Answer:
vₐ = v_c 
Explanation:
To calculate the escape velocity let's use the conservation of energy
starting point. On the surface of the planet
Em₀ = K + U = ½ m v_c² - G Mm / R
final point. At a very distant point
Em_f = U = - G Mm / R₂
energy is conserved
Em₀ = Em_f
½ m v_c² - G Mm / R = - G Mm / R₂
v_c² = 2 G M (1 /R - 1 /R₂)
if we consider the speed so that it reaches an infinite position R₂ = ∞
v_c = 
now indicates that the mass and radius of the planet changes slightly
M ’= M + ΔM = M ( 
)
R ’= R + ΔR = R ( 
)
we substitute
vₐ = 
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
vₐ = v_ c ( 
)
we make the product and keep the terms linear
vₐ = v_c
In both cases less energy is required
But comparetively Mg require more energy than K
Let's see the electron configuration of Both
- [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
- [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹
K has only one valence electron so very less ionization enthalpy so less energy required
Mg has 2 so more IE hence more energy required
The shades are very different
Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
