A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?
2 answers:
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a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:
where R is the radius of the wheel.
The period of revolution is:
Therefore, the tangential speed of the book is:
Answer:
a) 1321.45 N
b) 1321.45 N
c) 2.66 m/s^2
d) 2.21*10^-22 m/s^2
Explanation:
Hello!
First of all, we need to remember the gravitational law:
Were
G = 6.67428*10^-11 N(m/kg)^2
m1 and m2 are the masses of the objects
r is the distance between the objects.
In the present case
m1 = earth's mass = 5.9742*10^24 kg
m2 = 497 kg
r = 1.92 earth radii = 1.92 * (6378140 m) = 1.2246*10^7 m
Replacing all these values on the gravitational law, we get:
F = 1321.45 N
a) and b)
Both bodies will feel a force with the same magnitude 1321.45 N but directed in opposite directions.
The acceleration can be calculated dividing the force by the mass of the object
c)
a_satellite = F/m_satellite = ( 1321.45 N)/(497 kg)
a_satellite = 2.66 m/s^2
d)
a_earth = F/earth's mass = (1321.45 N)/( 5.9742*10^24 kg)
a_earth = 2.21*10^-22 m/s^2