Our Sun emits most of its radiation at a wavelength of 550 nm. If a star were 3.50 times hotter than our Sun, it would emit most

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

Utilizing the acceleration formula,

Answer:

The overall length of the spiral, designated as L, is calculated to be 5378.01 m

Explanation:

Provided information:

Inner radius R1=2.5 cm

and outer radius R2= 5.8 cm.

The thickness of the spiral winding is (d) =1.6 \mu m = 1.6x 10^{-6} m

The total length of the spiral can be computed as

= 5378.01 m

The answer is 13900589.

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

diameter (d) = 3.2 m

mass (m) = 42 kg

angular speed (ω) = 4.27 rad/s

Using the conservation of energy principle, we have

mgh = 0.5 mv² + 0.5Iω²...equation 1

where

Inertia (I) = 0.5mr²

ω = v/r

Revising equation 1, it turns into

mgh = 0.5 mv² + 0.5(0.5mr²)(v/r)²

resulting in gh = 0.5 v² + 0.5(0.5)v²

This simplifies to 4gh = 2v² + v²

thus h = 3v² ÷ 4g... equation 2

Given ω = v/r, we find v = ωr = 4.27 × (3.2 ÷ 2)

which yields v = 6.8 m/s

Next, substituting the value of v into equation 2 gives us

h = 3v² ÷ 4g

h = 3 × (6.8)² ÷ (4 × 9.8)

h = 3.57 m