The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "
Doppler's effect."
The standard formula for the Doppler effect is:

-- (A)
Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.
Where,
g = Speed of sound = 340m/s.

= Velocity of the observer relative to the medium =?.

= Velocity of the source in relation to the medium = 0 m/s.

= Frequency emitted from the source = 400 Hz.

= Frequency recognized by the observer = 408 Hz.
Substituting the given values into equation (A) will yield:


Solving the above will result in,

= 6.8 m/s
The correct result = 6.8m/s
C) B ⃗ = A ⃗ + C ⃗ Pay attention to the direction of the arrows (vectors).
The infinitesimal charge dQ on a layer with thickness dr is expressed as
dQ = (charge density) × (surface area) × dr
dQ = ρ(r)4πr²dr
∫ dQ = ∫ (a/r)4πr²dr
∫ dQ = 4πa ∫ rdr
Q(r) = 2πar² - 2πa0²
Q = 2πar² (= total charge confined within a spherical surface of radius r)
According to Gauss's Law:
(Flux through surface) = (charge enclosed by surface)/ε۪
(Surface area of sphere) × E = Q/ε۪
4πr²E = 2πar²/ε۪
<span>E = a/2ε۪
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Response:
x = 1.63 m
Details:
mass (m) = 10 kg
μk = 0.3
velocity (v) = 3.1 m/s
Assuming that the weight of the computer is largely applied to the belt instantaneously, we can implement the constant acceleration equation below
x = 
where a = μk.g, thus
x =
/2μk.g
x = (3.1 x 3.1)/(2 x 0.3 x 9.8)
x = 1.63 m
Answer:
Speed = 0.296m/2
Period = 0.203 s
Explanation:
If by 'long' you're referring to the waves' wavelength, then the wavelength
.
The waves have a frequency of 14.8 cycles every 3 seconds, or
.
The interplay between the wavelength
, frequency
, and speed
of the waves is defined as:

We input the values
and
leading to:
To determine the period
, one simply calculates the inverse of the frequency, or

