(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
<h3>
Net force on the third child</h3>
Apply Newton's second law of motion;
∑F = ma
where;
- ∑F is net force
- m is mass of the third child
- a is acceleration of the third child
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
- the wagon
- the children outside the wagon
<h3>Free body diagram</h3>
→ → Ф ←
1st child friction wagon 2nd child
<h3>Acceleration of the child and wagon system</h3>
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
<h3>When the frictional force is 21 N</h3>
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
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<span>Velocity, you divide distance/time </span>
The best and most correct answer among the choices provided by your question is the second choice or letter B.
The researchers’ conclusion was not justified because t<span>he control group was not treated the same as the experimental group.</span>
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Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :
We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :
At t = 1 s,
Current,
So, the current at t = 1 s is 3 A.
For lowest current,
Hence, this is the required solution.
These are the Kepler's laws of planetary motion.
This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.
The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.
The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.
