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3 years ago

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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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

faltersainse [42]

<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

Displacement in 4 seconds = 332 - 76 = 256 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{256}{4}=64ft/s$

Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

Displacement in 4 seconds = 78 - 332 = -254 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{-254}{4}=-63.5ft/s$

Average velocity in second 4 seconds is 63.5 ft/s downward

3 0

3 years ago

In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

6 0

3 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

3 years ago

What exited kaveh pahlevan about his work?

USPshnik [31]

Answer:

i dont know

Explanation:

i dont know since you didn't provide something to base off of

7 0

3 years ago

A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

$q = 1.6 \times 10^{-19} C$

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

8 0

4 years ago