Being underwater REDUCES the ability to localize sound. This is because, the resonance frequency of the external ear is reduced when the outer ear canal is fill with water. The presence of the water - air interface when one is under water and the reduction in the impedance matching ability of the middle ear also contribute to the reduction in one's ability to localize sound underwater.
Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = 
and filling in:
Δx = 
and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
Answer:
a) V = k 2π σ (√(b² + x²) - √ (a² + x²))
,
b) E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
Explanation:
a) The expression for the electric potential is
V = k ∫ dq / r
For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P
R = √(x² + r²)
The charge on a ring is
σ = dq / dA
The area of a ring is
A = π r
dA = 2π r dr
So the charge is
dq = σ 2π r dr
We substitute
V = k σ 2pi ∫ r dr / √(r² + x²)
We integrate
V = k 2π σ √(r² + x²)
We evaluate from the lower limit r = a to the upper limit r = b
V = k 2π σ (√(b² + x²) - √ (a² + x²))
b) the electric field and the potential are related
E = - dV / dx
E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))
E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
