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
Answer:
Buffalo, NY
Explanation:
Temperature in Buffalo, NY = -29°C
In order to compare the temperatures we need to convert them to the same scale.
So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.
Hence, it was colder in Buffalo, NY than in Anchorage, AL.
Answer:
X: Low potential energy
Y: High Potential energy
Z: Flow of electrons
Explanation: From the figure, it's obvious that Z is the flow of electrons, as shown by the arrow demonstrating the direction of the flow. Because of this, we can easily nullify choices B and C.
From the figure, we can notice that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.
The amplitude of a wave tells us about the intensity or brightness of the light relative to other light waves of the same wavelength.
Answer:
unknow e and f
Explanation:
In experiments with alpha particles that are obtained by the method of radioactive decay of atoms, some parameters are known
a) Known. The initial velocity is given by the energy of the particles entities by the atomic nuclei
b) Known. The particle charge always 2e, helium core
c) Known. It is set in the given experiment, in general it is selected as zero
d) Known. Placed by the experimenter
e) Unknown. The speed depends on the interactions with the system
f) Unknown. It depends on the interactions with the system, because the position depends on the interactions
g) Known. It is always the value of a helium nucleo
