It's mass and acceleration
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
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We can conclude that star A is closer to us than star B.
In fact, the absolute magnitude gives a measure of the brightness of the star, if all the stars are placed at the same distance from Earth. So, it's a measure of the absolute luminosity of the star, indipendently from its distance from us: since the two stars have same absolute magnitude, it means that if they were at same distance from Earth, they would appear with same luminosity. Instead, we see star A brighter than star B, and the only explanation is that star A is closer to Earth than star B (the closer the star A, the brigther it is)
I believe that the statement above is TRUE. I can say that it's true because based on the definition of what reciprocal liking is, this is the probability of having to like someone who also likes them in return. Therefore, if they like each other, this means that there is a low risk of rejection in the relationship.