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LenKa [72]
3 years ago
11

The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c

alculate the time taken for the echo of the sound to return to the bat.
Physics
1 answer:
Bingel [31]3 years ago
4 0

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

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A thin disk of mass 2.2 kg and radius 61.2 cm is suspended by a horizonal axis perpendicular to the disk through its rim. The di
Alinara [238K]

Answer: The period of the subsequent simple harmonic motion is 1.004 sec.

Explanation:

The given data is as follows.

  Mass of disk (m) = 2.2 kg,     radius of the disk (r) = 61.2 cm,

Formula to calculate the moment of inertia around the center of mass is as follows.

          I_{cm} = \frac{1}{2}mr^{2}  

                     = \frac{1}{2} \times 2.2 kg \times (61.2)^{2}

                     = 0.412 kg m^{2}

Also,

Distance between center of mass and axis of rotation (d) = r = 0.612 m

Moment of inertia about the axis of rotation (I)

        I = I_{cm} + md^{2}

       I = 0.412 + (2.2) \times (0.612)^{2}

          = 0.339 kgm^{2}

Now, we will calculate the time period as follows.

        T = 2\pi sqrt{\frac{I}{mgd})

        T = 2 \times 3.14 sqrt{(\frac{0.339}{2.2 \times 9.8 \times 0.612}

        T = 1.435 sec

      T = 2 \times 3.14 \times 0.16

          = 1.004 sec

Thus, we can conclude that the period of the subsequent simple harmonic motion is 1.004 sec.

7 0
3 years ago
what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this
natali 33 [55]

Answer:

55 miles / hour = 1.475232 kilometers / minute

6 0
3 years ago
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An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 7570 rpm while it is be
anastassius [24]

Answer:

The magnitude of the average angular acceleration is calculated as 1822.36\ rad/s^{2}

Explanation:

Maximum speed that can be attained by the disk, N_{m} = 10,000 rpm

Speed of spinning of the disk, N = 7570 rpm

Time taken to come to rest, t = 0.435 s

Now,

The initial angular velocity is given by:

\omega = \frac{2\pi N}{60} = 792.73\ rads^{-1}

Final angular velocity, \omega' = 0\ rads^{- 1}

The average angular acceleration of the disk can be computed by using the kinematic eqn:

\omega' = \omega + \alpha t

0 = 792.73 + 0.435\alpha

\alpha = - 1822.36\ rads^{- 2}

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a) "gravitation" is the force causing you to go down a waterslide

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d) fluid friction being the weakest friction, switching to sliding friction means a higher decrease in speed and therefore removing the water from a slide will decrease our speed

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