Ask question

Ask question

All categories

3 years ago

14

If the bat emits a sound at a frequency of 80.2 kHzkHz and hears it reflected at a frequency of 83.5 kHzkHz while traveling at a

speed of 3.9 m/sm/s , calculate the speed of the insect.

1 answer:

solong [7]3 years ago

8 0

To solve this problem we will apply the concepts related to the Doppler Effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other.

Mathematically it can be described as

$v' = v (\frac{f' ( v - v_{bat}) - f ( v + v_{bat})}{f'(v-v_{bat}) + f ( v + v_{bat})})$

Where,

v = Speed of sound

f' = Reflected frequency

f = Emitted Frequency

$v_{bat}$ = Speed of the object

Assuming that the speed of the medium is the speed of sound of 343m / s, we will then have that:

$v' = 343 (\frac{(83.5) (343 -3.9) - 80.2(343+3.9)}{83.5(343-3.9) + 80.2(343+4.3)})$

$v' = 3.0134m/s$

You might be interested in

Please answer any of these thanks !

KIM [24]

1). The equation is: (speed) = (frequency) x (wavelength)

Speed = (256 Hz) x (1.3 m) = 332.8 meters per second

2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.

If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.

If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.

3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)

Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.

That's ' 200 k Hz ' .

Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !

7 0

3 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

Where is hydrogen grouped on the periodic table?

k0ka [10]

Answer:

I think it is D but don't count on it

4 0

3 years ago

Read 2 more answers

PLEASE HELP! It’s urgent... and please show your work!!

bagirrra123 [75]

Answer:

A) 3.8 x 10

Explanation:

4 0

3 years ago