(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
Learn more about net force here: brainly.com/question/14361879
#SPJ1
Democritus was the one who did not have experimental evidence to support his theory of the atom.
Answer: Option 4
<u>Explanation:
</u>
The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.
After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.
This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.
Answer:
B) 4500 Pa
Explanation:
As pressure is force per unit area,
P = F/A
It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.
The possible areas are
0.30(0.40) = 0.12 m²
0.30(0.50) = 0.15 m²
0.40(0.50) = 0.20 m²
The pressure when the face with the largest area (0.20 m²) is down is
P = 900 / 0.20 = 4500 N/m² or 4500 Pa
the other possible pressures would be
900/0.15 = 6000 Pa
900/0.12 = 7500 Pa
which are both larger than our solution.
Answer:
E
Explanation:
Using Coulomb's law equation
Force of the charge = k qQ /d²
and E = F/ q
substitute for F
E = ( K Qq/ d² ) / q
q cancel q
E = KQ / d²
so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²
Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.
That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.
