Meteorite On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, creating a dent about 22 cm deep. If the initial speed of the meteorite was 550 m/s, what was the average force exerted on the meteorite by the car?

8427375 N

Explanation:

From newton's equation of motion,

F = ma Equation 1

Where F = average force exerted on the meteorite by the car, m = mass of the meteorite, a = acceleration.

We can find the value of acceleration from newton's equation of motion

using,

v² = u²+2as Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

Make a the subject of the equation

a = (v²-u²)/2s Equation 3

Note: When the meteorite struck the car, it was brought to rest

Given: u = 550 m/s, v = 0 m/s ( brought to rest), s = 22 cm = 0.22 m

Substitute into equation 3

a = (0²-550²)/(2×0.22)

a = -302500/0.44

a = -687500 m/s²

Also given: m = 27 pound = (27×0.454) = 12.258 kg

Substitute into equation 1

F = 12.258(-687500)

F = -8427375 N

Note: The negative sign tells that the force act in opposite direction to the motion of the meteorite

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

higher potential

explanation:

because it is in a high speed so it must move into a of high potential