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BaLLatris [955]
3 years ago
7

a whistle you use to call your hunting dog has a frequency of 21 khz, but your dog is ignoring it. you suspect the whistle may n

ot be working, but you can't hear sounds above 20 khz. to test it, you ask a friend to blow the whistle, then you hop on your bicycle.At what minimum speed should you ride to know if the whistle is working?
Physics
1 answer:
laila [671]3 years ago
4 0

To solve this problem we will apply the concepts related to the Doppler effect. According to this concept, it is understood as the increase or decrease of the frequency of a sound wave when the source that produces it and the person who captures it move away from each other or approach each other. Mathematically this can be described as

f = f_0 (\frac{v-v_0}{v})

Here,

f_0 = Original frequency

v_0 = Velocity of the observer

v = Velocity of the speed

Our values are,

v = 340m/s \rightarrow \text{Speed of sound}

f = 20kHz \rightarrow \text{Apparent frequency}

f_0 = 21kHz \rightarrow \text{Original frequency}

Using the previous equation,

f = f_0 (\frac{v-v_0}{v})

Rearrange to find the velocity of the observer

v_0 =v (1-\frac{f}{f_0})

Replacing we have that

v_0= (340m/s)(1-\frac{20kHz}{21kHz})

v_0 = 16.19m/s

Therefore the velocity of the observer is 16.2m/s

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Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t
laila [671]

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

h(t)=-16t^{2}+64t+112

Differentiate with respect to t on both the sides, we get

\frac{dh}{dt}=-32t+64

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

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elena-s [515]

Answer:

5 Miles

Explanation:

5miles is the answer

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