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il63 [147K]
3 years ago
12

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l

ow frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 cm apart, and the speed of sound generated is 340 m/s. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear
Physics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

\Delta t = \frac{0.18\ m}{340\ m/s}

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

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What is the formula for conservation of momentum
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Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

Explanation:

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2 years ago
What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?
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Answer:

Explanation:

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The following will tend to increase the magnetic force acting on the rotor in an induction motor.

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3 years ago
When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.
algol [13]

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The body has negative acceleration PR a deceleration.

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HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0
2 years ago
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
muminat

Answer:

4°C

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6 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
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