The path of a meteor passing Earth is affected by its gravitational force and falls to Earth's surface. Another meteor of the sa
2 answers:
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Answer:
At this position, the magnetic field equals ZERO
Explanation:
The magnetic field produced by a moving charge is described as
Here, we determine the direction of the magnetic field using
Thus, we find
Leading to a magnetic field of ZERO
Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent
Answer:
Explanation:
We start with the fact that
Initially, the spacecraft was at rest, u = 0The final velocity of the rocket is given as v = 11 m/sThe distance that the rocket covers during acceleration is given as d = 213 mWe seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying thethird motion equation we can find the acceleration.
Thus, the acceleration felt by those inside the spacecraft is .
a)
i) 120 s
ii) 1.57 m/s
b)
i) Refer to the attached diagram
ii) Up
c)
d) Greater than
Explanation:
The problem does not provide full details: consult the attachments for the complete text.
a)
The revolution period of the book equals the total duration needed for the book to make one full revolution.
By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.
We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.
This results in the revolution period being
T = 210 - 90 = 120 s
ii)
The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.
Mathematically:
where
R represents the wheel radius
T = 120 s indicates the period
Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as
d = +30 - (-30) = 60 m
This means the radius calculates to
R = d/2 = 30 m
So, the final speed is
b)
i) Please consult the attached free-body diagram for the book when at its lowest point.
Two forces act on the book at the lowest position:
- The weight of the book, represented as
where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.
- The normal force the bench exerts on the book is represented by N. This force acts upward.
ii)
While at its lowest position, the book maintains a horizontal motion at constant speed.
Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.
According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:
where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.
c)
As discussed in part b), there are two forces influencing the book at the lowest point:
- The weight, , directed downward
- The normal force from the bench, N, directed upward
Given that the book is in uniform circular motion, the net force must match the centripetal force , leading us to the equation:
where
represents the speed of the book
R stands for the radius of the circular path.
We derive an expression for the normal force:
d)
As per the discussions in parts c) and d):
- The normal force acting on the book at its lowest point becomes
- The weight (gravitational force) of the book is
Upon comparing these two equations, we conclude:
Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.
