Which form of energy is due to an object's position or location
2 answers:
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Answer:
hello your question is incomplete attached below is the missing part
answer : short period oscillations frequency = 0.063 rad / sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
Explanation:
first we have to state the general form of the equation
=
where :
comparing the general form with the given equation
= 18.2329
hence the short period oscillation frequency ( ) = 0.063 rad/sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
Answer:
Part a)
Part b)
Explanation:
Part a)
If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
Part b)
If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m