1) The electric potential at a distance r from a single point charge is given by
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is
So the potential at distance
is
2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
Answer:
8m/s^2
Explanation:
<em>Acceleration</em><em>=</em><em>initial </em><em>velocity</em><em>-final </em><em>velocity</em><em>/</em><em>time</em>
<em>initial</em><em> </em><em>velocity(</em><em>u)</em><em> </em><em>is </em><em>3</em><em>2</em><em> </em><em>final </em><em>velocity</em><em> (</em><em>v)</em><em> </em><em>is </em><em>9</em><em>6</em>
<em>therefore</em>
<em>a=</em><em>9</em><em>6</em><em>-</em><em>3</em><em>2</em><em>/</em><em>8</em>
<em>a=</em><em>8</em><em>m</em><em>/</em><em>s </em><em>squared</em>
<em>hope </em><em>it </em><em>helps</em>
<span>Young modulus E = stress/strain where E is the elastic modulus. We seek to calculate the the strain.
But we would have to get the stress first. Tensile stress = Force/Area. Force = 20100N. And the Area = 9.54* 13.9 = 132.606Nm^2.
Strain = stress/E.where E = 79. Hence we have 132.606/79 = 1.67</span>
Answer:
Explanation:
So we want the speed to go from 25 m/s to 0 m/s in 170 m, but the time needs to incorporate the reaction time, so the slowing down will not start until .68 s pass. Or, in other words, the train will travel an extra 25 m/s * .68 s = 17 m. This means, instead of 170 m to slow down it has 153. Hopefully that makes sense. With this information we can use the equation vf^2-vi^2=2ad. If that equation is unfamiliar you need to get a better handle on your physics equations.
Anyway, let's plug in.
vf = 0 m/s
vi = 25 m/s
a is what we're trying to find
d = 153 m
vf^2-vi^2=2ad
a = (vf^2-vi^2)/(2d)
Can you handle figuring it out from there? or if there is something you don't understand let me know.
Well let's see:
A). No. A capacitor doesn't measure anything.
B). No. The power delivered to the circuit is determined by
the battery or power supply and all the things in
the circuit that dissipate energy. A capacitor doesn't
do any of these things.
C). No. If any current actually flows between its plates,
the capacitor is shot and can't do its job, and
must be replaced.
D). Yes. A capacitor stores charges on its plates, and
electrical energy in the field between its plates.