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3 years ago

A high frequency sound will have a ?

Physics

2 answers:

stepladder [879]3 years ago

7 0

Answer:

The frequency of a sound wave is what your ear understands as pitch. A higher frequency sound has a higher pitch and the lower the period

masya89 [10]3 years ago

6 0

Answer:

High-frequency sound waves produce high-pitched sounds, and low-frequency sound waves produce low-pitched sounds.

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Once again we have a skier on an inclined plane. The skier has mass M and starts from rest. Her speed at the bottom of the slope

mars1129 [50]

Answer:

v = 31.3 m / s

Explanation:

The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.

Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane

Highest point

Em₀ = U = m g y

Lowest point

$Em_{f}$ = K = ½ m v²

As there is no friction, mechanical energy is conserved

Em₀ = $Em_{f}$

m g y = ½ m v²

v = √ 2 g y

Where we can use trigonometry to find and

sin 30 = y / L

y = L sin 30

Let's replace

v = RA (2 g L sin 30)

Let's calculate

v = RA (2 9.8 100.0 sin30)

v = 31.3 m / s

4 0

3 years ago

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick

rewona [7]

Answer:

The new time period is $T_2 = 3.8 \ s$

Explanation:

From the question we are told that

The period of oscillation is $T = 5 \ s$

The new length is $l_2 = 0.76 \ m$

Let assume the original length was $l_1 = 1 m$

Generally the time period is mathematically represented as

$T = 2 \pi \sqrt{ \frac{ I }{ mgh } }$

Now I is the moment of inertia of the stick which is mathematically represented as

$I = \frac{m * l^2 }{12 }$

$T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }$

Looking at the above equation we see that

$T \ \ \ \alpha \ \ \ l$

=> $\frac{ T_2 }{T_1} = \frac{l_2}{l_1}$

=> $\frac{ T_2}{5} = \frac{0.76}{1}$

=> $T_2 = 3.8 \ s$

3 0

3 years ago

Select all correct answers....Covalent compounds

yaroslaw [1]

I know for sure that the third one is correct

5 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet

Leno4ka [110]

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

8 0

3 years ago