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3 years ago

If you were measuring a lower-pitched sound wave, you would measure _____.

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Answer: B

To measure a lower-pitched sound wave, you would measure fewer compressions per second. Pitch <span>relates to frequency by measuring how many times a second the particles vibrate. Therefore, the lower-pitched sounds, the fewer compression waves or sound waves are produce because low-frequency sounds have greater distance between each wave.</span>

Moreover, the frequency of a wave is measured as the number of complete back-and-forth vibrations of a particle of the medium per unit of time. <span>The universal unit for frequency used is the Hertz (abbreviated Hz), where 1 Hertz = 1 vibration/second. </span>

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A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₁² v₂

$v_1= \dfrac{r_1^2}{r_2^2} v_2$

$v_1= \dfrac{1.4^2}{2.8^2} v_2$

$v_1= 0.25 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)$

$\Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)$

$v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}$

$v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}$

v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)² x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0

4 years ago

2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)

il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0

1 year ago

What is the displacement of the GBS cross-country team if they begin at the school, run 10 miles, and finish back at the school?

olga2289 [7]

Answer:

Explanation:

8 0

3 years ago

"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

vagabundo [1.1K]

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0

3 years ago

The illustration shows a rollercoaster and indicates four different positions the car might be at as it moves along the track. A

Vanyuwa [196]

Answer:

Point a

Explanation:

The potential energy of an object is given by :

P = mgh

m is mass, g is acceleration due to gravity, h is height above ground level.

Potential energy is directly proportional to the position of an object.

In the attached figure, the maximum height is shown at point (a). It means it will have maximum potential energy at a as compared to b,c and d.

5 0

3 years ago