Answer:
a) 0.15 μC b) 9.4*10¹¹ electrons.
Explanation:
As the total charge must be conserved, the total charge on the spheres, after being brought to contact each other, and then separated, must be equal to the total charge present in the spheres prior to be put in contact:
Q = +8.2μC +9.0 μC +(-7.8 μC) + (-8.8 μC) = +0.6 μC
As the spheres are assumed perfect conductors, as they are identical, once in contact each other, the excess charge spreads evenly on each sphere, so the final charge, on each of them, is just the fourth part of the total charge:
Qs = Qt/4 = 0.6 μC / 4 = 0.15 μC.
b) As the charge has a positive sign, this means that each sphere has a defect of electrons.
In order to know how many electrons are absent in each sphere, we can divide the total charge by the charge of one electron, which is the elementary charge e, as follows:
Answer:
I₂ = I₀ 1/4
Explanation:
For this exercise we use the definition of intensity which is the power per unit area
I = P / A
The emitted power is constant, so
P = I A
We can write this equation for the start and end point with index 2
I₀ A₀ = I₂ A₂
I₂ = I₀ A₀ / A₂
The spot area is the area of the circle
A₀ = π r₀²
We substitute
I₂ = I₀ r₀² / r₂²
It indicates that the radius of the spot is twice the initial radius
r₂ = 2 r₀
I₂ = I₀ (r₀ / 2 r₀)²
I₂ = I₀ 1/4
Answer:
b) 6
Explanation:
Given
v(t)=3t²+6t
X(0) = 2
X(1) = ?
Knowing that
v(t)=3t²+6t = dX/dt
⇒ ∫dX = ∫(3t²+6t)dt
⇒ X - X₀ = t³ + 3t²
⇒ X(t) = X₀ + t³ + 3t²
If X(0) = 2
⇒ X(0) = X₀ + (0)³ + 3(0)² = 2
⇒ X₀ = 2
then we have
X(t) = t³ + 3t² + 2
when
t = 1
X(1) = (1)³ + 3(1)² + 2
X(1) = 6
Answer:
Up And Down
Explanation:
In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.
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