On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite’s mass was about
4×106kg; its speed was about 15 km/s. Had it entered the atmosphere vertically, it would have hit Earth’s surface with about the same speed.
(a) Calculate the meteorite’s loss of kinetic energy (in joules) that would have been associated with the vertical impact.
(b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is
4.2×1015J.
(c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

A) Loss in Kinetic Energy |ΔK| = 4.5 x 10^(14)J

B) Loss in kinetic energy in expression of megaton = 0.1 megaton

C) Number of Hiroshima bombs that are equivalent to the meteorite impact is 8 bombs

Explanation:

From the question,

Mass of meteorite(m) = 4 x 10^(6)kg

Initial speed = 15km/s or 15000m/s

A) If we assume that the meteorite had penetrated the earths surface until it stopped moving, it's final speed will be zero.

Change in kinetic energy from initial motion to final motion is;

ΔK = Kf - Ki

Where Kf is final kinetic energy and Ki is initial kinetic energy.

Formula for kinetic energy is K= (1/2)(mv^(2)).

So final velocity = 0m/s while initial velocity is 15000m/s

Thus,

ΔK = [(1/2)(4 x 10^(6)(0)^(2))] - [(1/2)(4 x 10^(6))((15000)^(2))]

= 0 - 4.5 x 10^(14)

= -4.5 x 10^(14)J

When we have negative value like this, we take the absolute value which is |ΔK| = |-4.5 x 10^(14)|

Thus, |ΔK| = 4.5 x 10^(14)J

B) from the question, 1 megaton of TNT, contains 4.2×10^(15) J of energy.

Thus to express the loss in kinetic energy as a multiple of TNT, we have ;

|ΔK| = [ (4.5 x 10^(14)J) x (1 x megaton of TNT)] /(4.2×10^(15))

= 0.1 megaton of TNT

C) Converting 13kilotons to megaton = 13000 x 10^(-6) = 13 x 10^(-3) megaton

Thus; Since each bomb contains 0.013 megaton, the number of bombs associated with the meteorite impact is =

[(0.1 megaton x 1 bomb)]/(0.013)= 0.1/0.013 = 7.69 which is approximately 8 bombs

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2