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shtirl [24]
3 years ago
12

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does n

ot change when the sound enters the water. The wavelength of the sound is 2.86 m in the air, and the temperature of both the air and the water is 20 degrees C. What is the wavelength in the water
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of  119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

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The instantaneous expression for the magnetic field intensity of a uniform plane wave propagating in the +y direction in air is
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3 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

\theta = 13674\ rev

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

5 0
3 years ago
An object starts from rest and has an acceleration given by a = 3.00 t^2 – 4.20 t. Determine how far the object is from its star
Korolek [52]

Answer:

38.7 units

Explanation:

a = 3.00 t² – 4.20 t

Integrate to find velocity as a function of time.

v = ∫ a dt

v = ∫ (3.00 t² – 4.20 t) dt

v = 1.00 t³ – 2.10 t² + C

The object starts at rest, so at t = 0, v = 0.

0 = 1.00 (0)³ – 2.10 (0)² + C

0 = C

v = 1.00 t³ – 2.10 t²

Integrate to get position as a function of time.

x = ∫ v dt

x = ∫ (1.00 t³ – 2.10 t²) dt

x = 0.250 t⁴ – 0.700 t³ + C

Find the difference in positions between t = 4.50 and t = 0.

Δx = [0.250 (4.50)⁴ – 0.700 (4.50)³ + C] – [0.250 (0)⁴ – 0.700 (0)³ + C]

Δx = 0.250 (4.50)⁴ – 0.700 (4.50)³

Δx = 38.7

The object moves 38.7 units.

7 0
3 years ago
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