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Stella [2.4K]
1 year ago
6

In its search for flying insects, a bat uses an echolocating system based on pulses of high frequency sound. These pulses are 2.

60 ms in duration, have a frequency of 50.0 kHz, and an intensity level of 120 dB at 1.25 m from the bat’s mouth. Assume the bat produces acoustic waves in a cone with a total angle of 32o . (a) What is the acoustic power of each pulse? (b) How much acoustic energy is there in each pulse? (c) What is the intensity level at a conical surface centered on the bat with radius equal to 6.50 m? (d) A June bug is located 6.5m from the bat. The effective cross-sectional area of the insect is 14.0 mm2 . How much of the acoustic power emitted by the bat strikes the insect? (e) Assume all the acoustic power intercepted by the June bug goes into the reflected wave which is hemispherical. What is the intensity level of the reflected wave at the bat’s ears?
Physics
1 answer:
oee [108]1 year ago
4 0

The amount of energy that a wave may transfer to a unit area of a surface each second is measured as the wave's intensity. Watts per square meter is a unit used to express intensity. A sound wave's frequency is equal to its rate of vibration, and its intensity is determined by its amplitude.

<h3>What is the intensity of sound waves affect the frequency?</h3>

The energy of a vibration is quantified in decibels as intensity or loudness (dB). A sound has a high intensity if it is loud.

Therefore, perceive noise as louder the higher the frequency, although frequency does not indicate how loud a sound is.

Learn more about sound waves here:

brainly.com/question/1585667

#SPJ1

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Consider the average speed of a runner who jogs around a track four times. The distance (400m) remains constant for each lap. Ho
Yuki888 [10]

D) decreases, inverse

Explanation:

The average speed of each lap decreases. This is an example of an inverse relationship.

Speed is the rate of change of distance with time. In calculating the average speed, we should understand that the total length of path is taken into consideration and so is the total time taken.

  Average speed = \frac{total distance}{total time}

It was explained that after every 5 seconds the speed of the runner becomes slower.

There is definitely an inverse relationship between average speed and the total time taken.

An inverse relationship is such that as one variable increases, the other diminishes.

As average speed reduces the time taken increases.

learn more:

Speed brainly.com/question/1386181

#learnwithBrainly

7 0
3 years ago
1.How does inertia affect a person who is not wearing a seatbelt during a collision?
JulsSmile [24]
It keeps them back on the seat so they don't fly out the windshield.
4 0
4 years ago
A jet plane acceleration along a straight run way from rest to a speed of 140 m/s in 50 seconds when it took off. Calculate:
labwork [276]

Answer:

Vf=140 m/s

Vi=0

t=50s

Explanation:

1. Acceleration is called change of velocity.

2. a=Vf-Vi/t

a=140-0/50

a=140/50

a=2.8m/s^2

7 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
mestny [16]
22.5 meters. I attached all my work below. I hope this helped. If anything is still unclear please feel free to comment back with any questions you still have! It is helpful to draw pictures for problems like these, but if you know what you're doing you don't have to.

3 0
3 years ago
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