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Answer:
please give me brainlist and follow
Explanation:
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;

Here, C is the capacitance,
is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.
Explanation:
It is given that,
Area of nickel wire, 
Resistance of the wire, R = 2.4 ohms
Initial value of magnetic field, 
Final magnetic field, 
Time, t = 1.12 s
Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :






Induced current in the loop of wire is given by :



So, the induced current in the loop of wire over this time is
. Hence, this is the required solution.
The answer is A.) igneous rocks