Answer:
A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust
Explanation:
why would use atmosphere to study the layers of earth? dont think thats possible
Answer:
a
x-component 
y-component 
b
Magnitude 
direction is 
Explanation:
From the question we are told that
The first vertical distance is 
The first horizontal distance is 
The second vertical distance is 
Generally the displacement is
x-component 
y-component 
Generally the helicopters displacement is mathematically evaluated as



The direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as
![\theta = tan ^{-1}[ \frac{48}{20}]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20tan%20%5E%7B-1%7D%5B%20%5Cfrac%7B48%7D%7B20%7D%5D)
=> ![\theta = tan ^{-1}[ 2.4 ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20tan%20%5E%7B-1%7D%5B%202.4%20%5D)
=> 
Answer:6.71 m/s
Explanation:
Given
Apple fall from a height of
We need to find the impact speed of apple which can be given by using
where v=final velocity
u=initial velocity
h=Displacement
Assuming initial velocity to be zero
substituting the value we get
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 ×
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 =
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 =
+ 1000 × 9.8 ×36
pressure at bottom P2 =4.52 ×
Pa
so from gas law

here p is pressure and v is volume and t is temperature
so put here value and find v1

V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
Answer:
A) U₀ = ϵ₀AV²/2d
B) U₁ = (ϵ₀AV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
C) U₂ = (kϵ₀AV²)/2d
Explanation:
A) The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U₀ = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
ϵ = kϵ₀
where k = dielectric constant
ϵ₀ = permissivity of free space/air/vacuum
A = Cross sectional Area of the capacitor
d = separation between the capacitor
If air/vacuum/free space are the dielectric constants,
So, k = 1 and ϵ = ϵ₀
U₀ = CV²/2
Substituting for C
U₀ = ϵ₀AV²/2d
B) Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵ₀AV²/2d₁
U₁ = ϵ₀AV²/(2(3d))
U₁ = (ϵ₀AV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
C) U₂ = CV²/2
Substituting for C
U₂ = ϵAV²/2d
The dielectric material has a dielectric constant of k
ϵ = kϵ₀
U₂ = (kϵ₀AV²)/2d