Discuss how energy conservation applies to a
1 answer:
Answer:
a) in the upper position. b) in the lower position. c) in the lower position. d) in the upper position. f) Its kinetic and potential energy will be 0, but the energy is transferred to the element or body that stopped the movement of the pendulum
Explanation:
In the attached image we have the sketch of a pendulum system.
A) The potential energy is maximum when the pendulum is in the upper position (image, fig 1) because the elevation (h) is maximum with respect to the reference point.
B) the potential energy is minimum when the pendulum is in the lower pasition (image, fig 2) because the elevation (h) is cero with respect to the reference point.
Note: When the pendulum is coming down the potential energy is transforming in kinetic energy.
C) The kinetic energy is maximum when the pendulum is in the lower position (image, fig 2), because the potential energy has been transformed in kinetic energy.
D) The kinetic energy is maximum when the pendulum is in the upper position (image, fig 1) because at this moment the pendulum is at rest it means its velocity is 0. We know that the kinetic energy depends on the velocity.
f) The energy is transferred to the element or body that stopped the movement of the pendulum
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Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × m/s
uniform electric field E = 1.18 × N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 × m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 ×
(t)
t = 11.80 × s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns