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AlladinOne [14]
3 years ago
8

What is the speed of a periodic wave disturbance that has a frequency of 4.82 Hz and a wavelength of 0.14 m?

Physics
1 answer:
Talja [164]3 years ago
8 0

Answer:0.6748 m/s

Explanation:

Frequency=4.82hz

Wavelength=0.14m

Speed=frequency x wavelength

Speed=4.82 x 0.14

Speed=0.6748

Speed=0.6748m/s

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A sound wave traveling at STP has a period of .0023 seconds. The wavelength of this sound is closest to
agasfer [191]

Answer: 3

Wavelength = 0.76m

Explanation:

When sound wave is travelling at standard temperature and pressure ( STP), its speed will be equal to normal speed of sound which is equal to 331 m/s

Given that the period of the wave is 0.0023 seconds.

Wave speed = wavelength/period

Substitute the wave speed and the period to obtain wavelength

331 = wavelength/0.0023

Cross multiply

Wavelength = 331 × 0.0023

Wavelength = 0.7613 metre

The wavelength of this sound is closest to 0.76 m

4 0
3 years ago
Pls answer this<br> quick as possible
mihalych1998 [28]

Answer:

The answer is the 4th one

Explanation:

Good luck man

7 0
3 years ago
A diagnostic sonogram produces a picture of internal organs by passing ultrasound through the tissue. In one application, it is
miss Akunina [59]

Answer:

1) \lambda = 11\times 10^{-4} m

2)2.51\times 10^9 Pa

Explanation:

Given data:

speed of sound v  = 1540 m/s

frequency f  = 1.40 MHz = 1.40 \times 10^6 Hz

density \rho = 1060 kg/m^3

1) we know that

v = f\lambda

\lambda = \frac[v}{f} = \frac{1540}{1.40\times 10^6}

\lambda = 11\times 10^{-4} m

2) we know that

v= \sqrt{\frac{modulus}{density}}

v^2 = \frac{\gamma}{\rho}

\gamma = v^2 \rho

\gamma = 1540^2 \times 1060 = 2.51 \times 10^9 Pa

6 0
4 years ago
A basketball player standing under the hoop launches the ball straight up with an initial velocity of v₀ = 3.25 m/s from 2.5 m a
matrenka [14]

Answer: 0.53m

Explanation:

According to the equation of motion v²= v₀²+2as

Since the body is launched upward, the final velocity at the maximum height will be "zero" since the body will momentarily be at rest at the maximum height i.e v = 0

Initial velocity given (v₀) = 3.25 m/s

The body is also under the influence of gravity but the acceleration due to gravity will be negative being an upward force (a = -g) and the distance (s) will serve as our maximum height (h)

The equation of motion will.now become

V = v₀² -2gh

Where v = 0 v₀ = 3.25m/s g = 10m/s h = ?

0 = 3.25² - 2(10)h

0 = 10.56 - 20h

-10.56 = -20h

h = 10.56/20

h = 0.53m

Therefore, the maximum height, h (in meters), above the launch point that the basketball will achieve is 0.53m

6 0
4 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0
3 years ago
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