Answer:
Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s
Answer:
The right answer is Democritus
Explanation:
He experienced bonding with his father which strongly resulted his interest in science and to the study of the atoms ;)
Your welcome
It depends on if it is a sound wave or light wave
Answer:
a) (-367231.63i , 367231.63i, 0) N/C
b) (0 , 0 , 367231.63i ) N/C
Explanation:
a)
Case x < -2.15

Case x > 2.15

Case -2.15 < x <+2.15

b)
Case x < -2.15

Case x > 2.15

Case -2.15 < x <+2.15

Answer:
a)
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo = 
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s