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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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Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
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The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
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If you decrease the length of a wire, what effect will it have on resistance
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The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}
tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

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Kinetic energy of the 5 kg object.

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Velocity of an object = v

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

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Given that, v = 2v'

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an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

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put values of k, q1 , h1 and h2 in equation 1


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similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

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