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2 years ago

The interference of two sound waves of slightly different frequencies produces

2 answers:

7 0

The interference produces 'beats' ... two new sounds
in addition to the original ones.

The frequencies of the beats are

-- the sum of the two original frequencies
and
-- the difference of the two original frequencies .

Technically, there are actually even more new sounds than that.
There are:
-- beats between the beats, and
-- beats between each beat and each of the original two sounds.
But their amplitudes are so low that they can be detected only with
sensitive equipment, and they can't be heard in the presence of
the two original sounds and the two 'first order' beats.

Leviafan [203]2 years ago

4 0

Beats.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.

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The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.

tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

$W=K_f -K_i$

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

$K_f =\frac{1}{2}mv^2$

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

$W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J$

5 0

3 years ago

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A spectrophotometer measures the transmittance or the absorbance. True or False

kicyunya [14]

Answer: FALSE

Explanation: Could you help me with a question?

5 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

katen-ka-za [31]

Answer:

a) F = 2.66 10⁴ N, b) h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

P₂ = P₀ = 1.01 105 Pa

inside

P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

P₁ = 0.85 1.01 10⁵ + 1000 9.8 5

P₁ = 85850 + 49000

P₁ = 1.3485 10⁵ Pa

the net force is

ΔP = P₁- P₂

Δp = 1.3485 10⁵ - 1.01 10⁵

ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

P = Fe / A

F = P A

the area of a circle is

A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

d = 100 cm (1 m / 100 cm) = 1 m

F = 3.385 104 pi / 4 (1) ²

F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

P₁ = P₂

0.85 P₀ + ρ g h = P₀

h = $\frac{P_o ( 1-0.850)}{\rho \ g}$

h = $\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}$

h = 1.55 m

4 0

2 years ago

The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons

bulgar [2K]

Answer: $\lambda =248.99 nm$

Explanation:

Given

Work function $\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98$

$h=6.626\times 10^{-34} J$

$c=2.998\times 10^8$

$\phi =\frac{hc}{\lambda }$

$\lambda =\frac{hc}{\phi }$

$\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}$

$\lambda =248.99 nm$

3 0

3 years ago