Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
Answer:
It would be. Quarter dark
Explanation:
Sun's lights going all the way to the moon but there will still a dark part of it
it's obviously won't be neither complete light nor completely dark
so the right answer is: Quarter dark
Answer:
10 N
Explanation:
Work done is the dot product of the force magnitude with distance and cosine of the angle betweenthem.
Equation to use: W=|F|*|d|*cosθ
Your unknown is the force F; You have one equation, oneunknown.
40j=(F)*(4m)*cos(0)
F=40j/4m
F=10N
Answer:
no lol.. is this actually for school?
