energy associate with position or shape
Answer:
a) [volts] = [N m / C],
b) The lines or surface that has the same potential are called equipotential
c) the equipotential lines must also be perpendicular to the electric field lines
Explanation:
a) find the units of the volt
the electric potential energy is
V = k q / r
V = [N m² / C²] C / m
V = [N m / C]
The electric potential is defined as
V = E .s
V = [N / C] [m]
V = [N m / C] = [volt]
we see that in the two expressions the same result is obtained therefore the volt is
[volts] = [N m / C]
b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.
c) The electric potential is defined as the gradient of the electric field
v =
therefore the equipotential lines must also be perpendicular to the electric field lines
<span>D. density is your answer</span>
Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,

And we know as well that

Replacing the aceleration value in the equation force we have,

Substituting our values we have,


The weight of the person is then,


<em>We can conclude that force on the ball is large to lift the ball</em>
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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