<h2>Answer: Neptune
</h2>
The dwarf planet Pluton, <u>has the most eccentric orbit</u> (more elliptical and elongated) of all the planets, and as a consequence its orbit is "intersected" by the orbit of Neptune.
However, <u>despite this intersection, there is no collision risk between these two bodies</u>, since the orbit of Pluto is located in an orbital plane different from that of the other planets and therefore different from that of Neptune, its nearest neighbor. In addition, the orbit of Pluton is inclined 
on the the ecliptic plane (plane where the other planets move around the Sun).
The distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.
<h3>What is distance?</h3>
Distance can be defined as the horizontal length between two points.
To calculate the distance at which the string must be attached to balance the rod, we use the formula below.
Formula:
- m₁gx = m₂g(1.2-x)................. Equation 1
From the question,
Given
- m₁ = 2.0 kg
- m₂ = 2.5 kg
- x = Distance of the string from the 2.0 kg mass
- g = Acceleration due to gravity.= 9.8 m/s²
Substitute these values into equation 1
- 2(9.8)(x) = 2.5(9.8)(1.2-x)
Solve for x
Collect like terms
- 19.6x+24.5x = 29.4
- 44.1x = 29.4
- x = 29.4/44.1
- x = 0.67 m
Hence, the distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.
Learn more about distance here: brainly.com/question/17273444
Answer:
a) L = 440 cm
Explanation:
In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are
λ₁ = 4L fundamental
λ₃ = 4L / 3 third harmonic
λ₅ = 4L / 5 five harmonic
The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are
λ₁ = 2L fundamental
λ₂ = 2L / 2 second harmonic
λ₃ = 2L / 3 third harmonic
λ₄= 2L / 4 fourth harmonic
They indicate that resonance occurs in the fourth harmonic, let's look for the frequency
v =λ f
for the fundamental
v = λ₀ f₀
V = 2L f₀
for the fourth harmonica
v = λ₄ f ’
v = L / 2 f'
2L f₀ = L / 2 f ’
f ’= 4 f₀
f ’= 4 440
f ’= 1760 Hz
for this frequency it has the resonance with the tube
f ’= 4L
L = f ’/ 4
L = 1760/4
L = 440 cm
b) let's find the frequency of the next harmonic in the tube
λ₃ = 4L / 3
λ₃ = 4 400/3
λ₃ = 586.6 cm
v = λf
f = v / λlam₃
f₃3 = 340 / 586.6
f3 = 0.579
as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances
<span>The minimum mass flow rate of coolant required to keep the coolant exit temp at or below 90 degrees C is at a rate of 35%. To find this you need to take the temperature times the power output to find this answer.</span>
