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Nastasia [14]
3 years ago
6

A hummingbird can flap its wings 200 Hz times per second. If the hummingbird produces waves that travel at 340 m/s by flapping i

ts wings, what is the wavelength of these waves?
Use Wave Speed Equation:

wave speed (in m/s) = frequency (in Hz) x wavelength (in m)
s = f
λ\lambda
λ
Physics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

1.7 m

Explanation:

s = f λ

340 m/s = (200 Hz) λ

λ = 1.7 m

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Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as
Reil [10]

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

5 0
3 years ago
A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of
DENIUS [597]

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × 10^{-3}  kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 ×10^{-2} m

to find out

the speed of the mass

solution

we will apply here conservation energy  that is

K.E + P.E = Total energy  ..................1

so that Total energy = K.E max = P.E max

we know  amplitude so we find out first P.E max that is  

PE max = K.E + P.E  

(1/2)kA² = (1/2)mv² + (1/2)kx²  

kA^² =  mv²+ kx²

so here v²  will be

v²  = k(A² - x²) / m  

v = √[(k/m)×(A² - x²)]  ............2

here x = (1/2)A   so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × 10^{-3} )×(3/4×(38.9 ×10^{-2})²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s

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