We have that the speed of a body covering a distance of 320 km in 4h is mathematically given as
V=22.22m/s is
<h3 /><h3>From the question we are told
calculate the speed of a body covering a distance of 320 km in 4h
Generally the equation for the Speed is mathematically given as
V=22.22m/s
Hence
The speed of a body covering a distance of 320 km in 4h is
V=22.22m/s
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Answer:
77.35 m / s
Ф = -17° from + X axis or 343° from + X axis
Explanation:
v1 = 75 m/s 25° east of north
v2 = 100 m/s 25° east of south
Write the velocities in vector form ,we get
Now add the velocity vectors to get the resultant of the velocities.
magnitude of resultant velocity is
= 77.35 m / s
The direction is Ф from X axis
Ф = -17° from + X axis or 343° from + X axis
Answer:
The force applied 275 N in a direction parallel to the hill
Explanation:
Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.
Let us decompose the weight into its two components, the angle T is taken from the axis and
Wx = W sin θ
Wy = W cos T
We write Newton's second law
∑ F = m a
X axis
The cyclist falls at a constant speed, which implies that the acceleration is zero
fr - W sin θ = 0
fr = mg sin θ
fr = 96 9.8 without 17
fr = 275 N
When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement . The force applied 275 N in a direction parallel to the hill
