Solutes
Explanation:
These cleaning agents and additives are examples of solutes.
- Solutes are substances that dissolves in other substances called solvents to form a solution.
- The solutes are the cleaning agents and anti-knock additives.
- The solvents are the gasoline.
- When solutes dissolves in solvents they form solutions.
- A solution is a mixture that is homogeneous.
- For a solute dissolve in a solvent, they must both have the same property.
- Like substances will only dissolve in like ones.
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Answer:
Explanation:
Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:
By using the 
The mean position is the position when <em>y</em> = 0, so:
rad/s
and 
in centimeters (cm).
Answer:
ratio = 1 : 4.5
Explanation:
If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:
The force F₂:
The ratio:
Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,
- Charge placed into the region,
where, 
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
Thus, the y-component of the electric force on this charge is
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
