The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this
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The question is incomplete. Here is the complete question.
Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed
. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed
, which is greater than
.
Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.
Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.
Answer: Part A:
Part B:
Explanation: First, let's write an equation of motion for each car.
Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:
where
is initial position
v is velocity
t is time
Car A started the race at a distance. So at t = 0, initial position is .
The equation will be:
Car B started at the starting line. So, its equation is
Part A: When they meet, both car are at "the same position":
Car B meet with Car A after units of time.
Part B: With the meeting time, we can determine the position they will be:
Since Car B started at the starting line, the distance Car B will be when it passes Car A is units of distance.
(a) If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.
(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.
The force on an object is determined by applying Newton's second law of motion;
F = ma
(a)
when the force is halved, F₂ = 0.5F₁,
mass is doubled, m₂ = 2m₁
Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.
(b)
when the force is doubled, F₂ = 2F₁,
mass is halved, m₂ = 0.5m₁
Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.
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