<span>since sound travels using mechanical waves and needs a material medium to propagate and since mechanical waves spread through vibrations ...and since hard materials have their atoms packed closely....they need to vibrate with a smaller amplitude to pass on the wave....thus sound travels faster in a denser medium than a less dense one.</span>
(a) 
According to Newton's second law, the force experienced by each balloon is given by:
F = ma
where
m = 0.021 kg is the mass
a = 1.1 m/s^2 is the acceleration
Substituting, we found:

The electrostatic force between the two balloons can be also written as

where
k is the Coulomb's constant
Q is the charge on each balloon
r = 16 m is their separation
Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

(b)
electrons
The magnitude of the charge of one electron is

While the magnitude of the charge on one balloon is

This charge can be written as

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

<span>NASA and the Mad Science Group of Montreal, Canada, have teamed in an effort to spark the imagination of children, encouraging more youth to pursue careers in science, technology, engineering and math. The two organizations recently signed a Space Act Agreement, officially launching the development of the Academy of Future Space Explorers.</span>
Answer:
Hey, bro here is the explanation....
Explanation:
Hope it helps...
Answer:
P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,
Explanation:
This problem of fluid mechanics let's start with the continuity equation to find the speed of water output
Q = A v
v = Q / A
The area of a circle is
A = π r² = π d² / 4
Let's look at the speeds at each point
v₁ = Q / A₁ = Q 4 /π d₁²
v₁ = 10 4 /π 0.5²
v₁ = 50.93 m / s
v₂ = Q / A₂
v₂ = 10 4 /π 0.25²
v₂ = 203.72 m / s
Now we can use Bernoulli's equation in the colon
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear
P₁ = P2 + ½ rho (v₂² - v₁²)
P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)
P₁ = 1.013 10⁵ + 2.205 10⁷
P₁ = 2.215 10⁷ Pa
la definicion de presion es
P₁ = F₁/A₁
F₁ = P₁ A₁
F₁ = 2.215 10⁷ pi d₁²/4
F₁ = 2.215 10⁷ pi 0.5²/4
F₁ = 4.3 106 N