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notsponge [240]
3 years ago
13

What is the highest pHoney bees beat their wings, making a buzzing sound at a frequency of 2.3 × 102 hertz. What is the period o

f a bee's wing beat?oint on a wave called?
Physics
1 answer:
VMariaS [17]3 years ago
5 0

The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

T=\frac{1}{f}

where

T is the period

f is the frequency

For the bee in this problem, the frequency of the sound wave emitted by it is

f=2.3 \cdot 10^2 Hz = 230 Hz

Therefore, the period of the sound wave is

T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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Light will travel more slowly in a material with a higher index of refraction 
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3 years ago
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
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Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

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6 0
3 years ago
A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
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Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

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f = \frac{0.40}{2} = 0.20 m

Now,

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From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

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the correct answer is 27 hours per week :) hope this helps


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