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Strike441 [17]
3 years ago
14

Find the wavelength of the ultrasonic wave emitted by a bat if it has a frequency of 4.0 * 10^4 Hz.

Physics
2 answers:
Oliga [24]3 years ago
6 0

ultrasonic wave is a type of sound wave. speed of sound in air at room temp is 343m/s.

speed = wavelength * frequency

wavelength = speed / frequency = 343/4*10^4

= 0.00858m

= 8.58mm



morpeh [17]3 years ago
5 0

Remark

You have to pick a velocity for an ultra sound wave. I'm going to take the sound velocity to be 331 m/s in air, because I'm assuming that is what is meant: it is the sound a bat would make once he emits the ultrasound to a microphone like device that receives it. The media through which that happens is air.

Givens

v = 331m/s

f = 4 *10^4 hz

Formula

velocity = wavelength * frequency

v = \lambda*f

331/(4*10^4) =  \lambda

0.008275 =  \lambda

8.275*10^-3 =  \lambda

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A ringing bell sends sound waves in all<br> directions<br> places<br> sides
Strike441 [17]

Answer:

direction

Explanation:

because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

4 0
3 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
elena-14-01-66 [18.8K]

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

  • c is speed of light
  • E₀ is the peak electric field
  • B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

8 0
2 years ago
Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?
tester [92]

Answer: 40\ N/C

Explanation:

Given

Magnitude of charge is q=0.5\ C

Force experienced is F=20\ N

Electric field intensity is the electrostatic force per unit charge

\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C

Thus, the electric field intensity is 40\ N/C

6 0
3 years ago
An imaginary line perpendicular to a reflecting surface is called ____refrac_____.
umka2103 [35]
Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)
4 0
3 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
3 years ago
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