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Strike441 [17]
2 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]2 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]2 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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You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon
Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

u_{1} = 0m/s

t_{1} = t

u_{2} = 43.12m/s

t_{2} = (t-2.2)s

The distance by the first balloon is

D = u_{1} t_{1}  + \frac{1}{2} at_{1}^2

where

a = 9.8m/s2

Inputting the values

D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2

The distance traveled by the second balloon

D = u_{2} t_{2}  + \frac{1}{2} at_{2}^2

Inputting the values

D = (43.12)(t-2.2)  + \frac{1}{2} (9.8)(t-2.2)^2

simplifying

D = 4.9t^2 + 21.56t -71.148

Substituting D of the first balloon into the D of the second balloon and solving

4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

D = 4.9(3.3)^2\\ D = 53.361 m

7 0
2 years ago
A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage
Alex787 [66]
The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip. 

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.
4 0
2 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
mina [271]

Answer:

n = 2.06 moles

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Pressure read + Zero Gauge pressure

Zero Gauge pressure = 14.7 psi

Pressure read = 480 psi

Total pressure = 480 psi + 14.7 psi = 494.7 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 33.66 atm

Temperature = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

T = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

33.66 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 2.06 moles

7 0
2 years ago
A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Nina [5.8K]

Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

3 0
2 years ago
A student hears a police siren. What would change the frequency that the student hears? Check all that apply.
ser-zykov [4K]
<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source. 

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>
4 0
2 years ago
Read 2 more answers
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