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3 years ago

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Imagine a sound wave with a frequency of 1 kHz propagating with a speed of 1500 m/s in water. Determine the phase difference in

radians between any two points on the wave separated by 5 cm.

1 answer:

spin [16.1K]3 years ago

6 0

Answer:

Phase difference will be equal to $\frac{\pi }{15}radian$

Explanation:

We have given frequency f = 1 kHz

Velocity of sound wave v = 1500 m/sec

It is given that wave are separated by 5 cm

So d = 0.05 m

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency

So $\lambda =\frac{v}{f}=\frac{1500}{1000}=1.5m$

$\beta$ is equal to $\beta =\frac{2\pi }{\lambda }=\frac{2\times 3.14}{1.5}=\frac{4\pi }{3}rad/m$

Phase difference is equal to $\Phi =\beta d=\frac{4\pi }{3}\times 0.05=\frac{\pi }{15}radian$

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anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

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3 years ago

Lab: newton's laws of motion assignment: lab report

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Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

<h3>What are Newton's motion laws?</h3>

Newton's motion laws are a set of scientific statements aimed at explaining the physical property of movement.

These laws explain why objects in movement tend to maintain the same velocity for a short period of time.

In conclusion, Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

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Answer:

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Explanation:

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T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

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The West is known for "wide, open spaces", cattle, mines, and mountains.

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A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

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Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

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As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

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Also, we know that

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or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

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thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

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3 years ago