Answer:
3.95 m
Explanation:
m = 1 kg, h = 100 m, k = 125 N/m
Let the spring is compressed by y.
Use the conservation of energy
potential energy of the mass is equal to the energy stored in the spring
m x g x h = 1/2 x ky^2
1 x 9.8 x 100 = 0.5 x 125 x y^2
y^2 = 15.68
y = 3.95 m
Answer:
<em>likely to decrease downstream in arid regions and increase downstream in temperate regions</em>
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Explanation:
Arid regions are is a region with a severe lack of water, usually to the extent that affect the organisms living in the region. Arid regions are characterized by a very low depth of rainfall per year. Temperate region on the other hand experience more distinct seasonal change and wider temperature change. Temperate regions get a fairly large amount of rainfall per year.
In arid regions, the soil is very dry, and the rate of infiltration and percolation is high relative to the amount of rainfall available. The effect is that more water is infiltrated into the soil as you move downstream, leading to a decrease in the discharge of a stream as you move downstream. Most temperate region have soils that are usually saturated in the peak of the rainfall season, leading to a greater stream discharge as you move downstream.
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
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
