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2 years ago

Describe a sound wave. Discuss what determines pitch and volume.

Physics

1 answer:

LenaWriter [7]2 years ago

8 0

Answer:

The amplitude of a sound wave is a reflection of how much energy is carried, which contributes to the intensity of the sound. Intensity is measured in decibels and is perceived as sound volume. Thus, the volume is proportional to the amplitude of the sound wave. The frequency of a sound wave is perceived as pitch.

Explanation:

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The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N

guajiro [1.7K]

Answer:

$\frac{r_1}{r_2}=6.9$

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

$P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}$

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

$\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9$

4 0

2 years ago

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

$(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]$

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

$y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]$

With this time we can find the horizontal distance that runs the projectile.

$x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]$

5 0

2 years ago

The angular momentum of a flywheel having a rotational inertia of 0.200 kg·m2 about its central axis decreases from 3.80 to 0.40

babymother [125]

Explanation:

Below is an attachment containing the solution

3 0

2 years ago

What did ancient astronomers think areas of the moon called mares might be?

Bumek [7]

The ancient astronomers think areas of the moon called mares might be Seas.

Option D

<u>Explanation</u>:

The surface area of Earth's moon is dark, large, and is basaltic plains which are formed by ancient volcanic eruptions. They were dubbed as Maria, "ancient astronomers" who misunderstood them as actual seas. They are less reflective than highlands. Due to their iron-rich composition, they tend to appear dark from the naked eye. The Maria cover approximately about 16% of surface mostly on side that is visible from Earth. The few Maria on side that is too far are much smaller and residing mostly in very large craters. The ancient astronomers mistook the surface area as look like actual seas.

4 0

2 years ago

g Determine the magnitude of the electric field at the surface of a lead-208 nucleus, which contains 82 protons and 126 neutrons

Nookie1986 [14]

Answer:

Explanation:

Electric field at the surface of the the lead 208 = KQ/ R²

where K = 8.99 × 10⁹ Nm² /C²

Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C

V of the lead = 208 v of a proton assuming they both are sphere

4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton

R³ = 208 r³

R = ∛( 208 r³) = 5.92r

replace r with 1.20 x 10-15 m

R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m

E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C

4 0

2 years ago