<span>Hydrocarbons are molecules that contain only carbon and hydrogen.</span>
Due to carbon's unique bonding patterns, hydrocarbons can have single, double, or triple bonds between the carbon atoms.
The names of hydrocarbons with single bonds end in "-ane," those
with double bonds end in "-ene," and those with triple bonds end in
"-yne".
The bonding of hydrocarbons allows them to form rings or chains.
Answer:
4.29 millimeters
Explanation:
Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

The frequency of the waves emitted by this bat is:

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

Answer:
,Assume that the average volume of an adult human body is one-tenth
cubic meter (0.10 m) and that there are two billion (2.0 x 109)
adults in the world.
a. What would be the total volume of all the adults in the world?
b. Compute the length of one edge of a cubic container that has a
volume equal to the volume of all the adults in the world.
Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.
The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.
"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.
The real question is:
What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?
Acceleration of gravity is
G · M / R²
= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²
= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ m / s²
That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.
In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g 
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g 
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
How energy is conserved
Em₀ = 
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)