Answer:A wave has a wavelength λ, which is the distance between adjacent identical parts of the wave. The wave velocity and the wavelength are related to the wave's frequency and period by vw=λT or vw=fλ.
Explanation:
Answer:
a)V= 0.0827 m³
b)P=181.11 x 10² N/m²
Explanation:
Given that
m = 81.5 kg
Density ,ρ = 985 kg/m³
As we know that
Mass = Volume x Density
81.5 = V x 985
V= 0.0827 m³
The force exerted by weight = m g
F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)
Area ,A= 4.5 x 10⁻² m²
The Pressure P


P=181.11 x 10² N/m²
Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:

And the angle measure from the y-axis is:

So the answer is 245.45km in a direction 21.45° west of north from city A
Answer:
λ = 3 10⁻⁷ m, UV laser
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
let's use trigonometry
tan θ = y / L
as in this phenomenon the angles are small
tan θ =
= sin θ
sin θ = y / L
we substitute
a y / L = m λ
let's apply this equation to the initial data
a 0.04 / L = 1 600 10⁻⁹
a / L = 1.5 10⁻⁵
now they tell us that we change the laser and we have y = 0.04 m for m = 2
a 0.04 / L = 2 λ
a / L = 50 λ
we solve the two expression is
1.5 10⁻⁵ = 50 λ
λ = 1.5 10⁻⁵ / 50
λ = 3 10⁻⁷ m
UV laser
If both waves have the same wavelength, then the amplitude of
their sum could be anything between 1 cm and 9 cm, depending
on the phase angle between them.
If the waves have different wavelengths, then the resultant is a beat
with an amplitude of 9 cm.