The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3
1 answer:
Answer:
has units of distance
has units of distance over time
has units of distance over
has units of distance over
Explanation:
Since the expression for the distance is:
then:
has units of distance
has units of distance over time
has units of distance over
has units of distance over
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
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Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:
which in our case becomes;
and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N