Ask question

Ask question

All categories

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

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>

You might be interested in

Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration

ollegr [7]

Answer:

Explanation:

The mass of the car doesn't matter because On a flat curve the mass of the car does not affect the speed at which it can stay on the curve. You would need the mass if you were solving the the centripetal force acting on the car, but not the acceleration.

$a=\frac{v^2}{r}$ and filling in

$a=\frac{(8.5)^2}{48.0}$ and we need 2 significant digits in our answer. That means that

a = 1.5 m/sec²

8 0

3 years ago

A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0

IgorLugansk [536]

Answer:

Magnitude of the force is

$F_3 = 2.83$

direction of the force is given as

$\theta = 45 degree$ West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

here we know that first force is of magnitude 2 N towards east

$\vec F_1 = 2 \hat i N$

second force is also of 2.0 N due North

$\vec F_2 = 2 \hat j$

now from above equation

$2\hat i + 2\hat j + \vec F_3 = 0$

$\vec F_3 = -2\hat i - 2\hat j$

so magnitude of the force is given as

$F_3 = \sqrt{2^2 + 2^2}$

$F_3 = 2.83$

direction of the force is given as

$\theta = tan^{-1}\frac{F_y}{F_x}$

$\theta = tan^{-1}\frac{-2}{-2}$

$\theta = 45 degree$ West of South

3 0

3 years ago

Read 2 more answers

Which of the following are physical properties of nonmetals? a. they are brittle in the solid form b. most are gases at room tem

Black_prince [1.1K]

Option (D) is correct that all of the above are physical properties of non metals.

#$# THANK YOU #$#

6 0

3 years ago

Read 2 more answers

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

4 years ago

- The Grayprational Force<br>with which the earth aracts<br>а,

Fudgin [204]

Answer:

gravitational?

Explanation:

5 0

3 years ago