Answer:
formula: A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º. The period of a simple pendulum is T=2π√Lg T = 2 π L g , where L is the length of the string and g is the acceleration due to gravity.
We want to study the impact of a sledgehammer and a wall.
Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.
When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.
This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.
Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.
If you want to learn more, you can read:
brainly.com/question/13952508
Answer:
<h2>Case i) if
</h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if
</h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>
Explanation:
As we know that the impedance of the circuit is given as
when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination
so it is
now we have
Case i) if
So initially if the circuit is inductive in nature then its net impedance will decrease after this
Case ii) if
So initially if the circuit is capacitive in nature then its net impedance will increase after this
Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
