Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
The formula for the energy in a capacitor , u in terms of q and c is q²/2c
<h3>What is the energy of a capacitor?</h3>
The energy of a capacitor u = 1/2qv where
- q = charge on capacitor and
- v = voltage across capacitor.
<h3>What is the capacitance of a capacitor?</h3>
Also, the capacitance of a capacitor c = q/v where
- q = charge on capacitor and
- v = voltage across capacitor.
So, v = q/c
<h3>
The formula for energy of the capacitor in terms of q and c</h3>
Substituting v into u, we have
u = 1/2qv
= 1/2q(q/c)
= q²/2c
So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c
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Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.
But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left
There are several approaches. The most favourable one (in my opinion) is this one:
1. Asking a question
2. Doing a research (how to answer this question)
3. Creating a hypothesis (NOT a thesis!)
4. Experimenting (to prove the hypothesis)
5. Analysing results from the experiment
6. Writing a thesis