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yarga [219]
3 years ago
10

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.
Physics
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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A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an
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.

7 0
3 years ago
Lab: newton's laws of motion assignment: lab report
poizon [28]

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.

Learn more about Newton's motion laws here:

brainly.com/question/10454047

#SPJ1

8 0
1 year ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

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

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0
2 years ago
What makes west region different?
anzhelika [568]

Answer:

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

7 0
3 years ago
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

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

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

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

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

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

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

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

                        = \sqrt{36}

                        = 6 m/s

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

4 0
3 years ago
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