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Blababa [14]
3 years ago
6

4. A tankful of liquid has a volumeof 0.2m3. What is the volume in (a)lities (b) cm3 (c)ml​

Physics
1 answer:
yan [13]3 years ago
5 0

Explanation:

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Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
3 years ago
A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is
tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency = f = 562Hz

Speed of sound in air = v = 331m/s

The definition of wavelength is,

\lambda = \frac{v}{f}

Here,

v = Velocity

f = Frequency

Replacing,

\lambda = \frac{331m/s}{562Hz}

\lambda = 0.589m

Therefore the wavelength of that tone in air at standard conditions is 0.589m

3 0
4 years ago
How do I ship a pringle in the mail without it breaking? This is a physics project.
vekshin1
Assuming you want it to be as small and lightweight as possible :

Cut a solid box roughly twice as big as the pringle. Put the pringle inside the box, and fill the remaining space with cotton, that will cushion the impacts. Be sure to apply the mention <em>FRAGILE</em> to the box, so that they'll take care of it properly.
5 0
4 years ago
A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The b
ch4aika [34]

Answer:

1.77 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............................ Equation 1

Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

make V the subject of the equation

V = (mu+m'u')/(m+m').................. Equation 2

Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)

Substitute into equation 2

V = (0.04×310+6.96×0)/(0.04+6.96)

V = 12.4/7

V = 1.77 m/s

Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

6 0
4 years ago
what did classical physics predict about electron flow as a result of light shining on a metal surface?
stiv31 [10]
This looks like the photo electric effect ... classical physics reckoned that if you shone an intense enough light beam on a metal you could get electrons ejected from the metal (maybe in analogy to thermionic emission - heat). It sort of "forgot" about the frequency and photon/particle nature of light.
Enter the "photo electric" effect experiment, Einstein's explanation, and the Nobel committee having an excuse to award E a Nobel prize, even though said prize was probably more for relativity.
8 0
3 years ago
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