Which is an advantage of storing data digitally?

a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one

prohojiy [21]

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be $37^o$ C is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

$f=\frac{v_T}{4L}$

where, $v_T=v\sqrt{\frac{T}{273} }$

Since, T=37+273 K = 310 K

v = 331 m/s

$\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s$

Using this, we get:

$f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz$

Hence, the fundamental frequency is 367.42 Hz.

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7 0

1 year ago

A person walks 60 meters in 1 minute. Then walks 30 meters in 2 minutes. What is thier average speed over the 3 minute walk?

Gennadij [26K]

Your answer would be 75 meters

8 0

3 years ago

Read 2 more answers

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago

A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res

Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0

3 years ago

A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

4 0

3 years ago