All categories

3 years ago

How do I find the speed of sound when given Frequency and displacement?

1 answer:

3 0

Answer: Taking into account sound is a wave, we can use the information of the displacement (generally given as a graph) to find the wavelength and frequency, then we can calculate the speed with the formula of the speed of a wave.

Explanation:

If we have the displacement graph of the sound wave, we can find its amplitude, its wavelength and period (which is the inverse of frequency).

Now, if we additionally have the frequency as data, we can use the equation of the speed of a wave:

$s=\lambda f$

Where:

$s$ is the speed of the sound wave

$\lambda$ is the wavelength

$f$ is the frequency

You might be interested in

You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,

aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

$P_{r}=\dfrac{I}{c}$

$P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}$

$P_{r}=2.75\ N/m^2$

We need to calculate the force

Using formula of force

$F=P\times A$

$F=P\times \pi r^2$

Put the value into the formula

$F=2.75\times\pi (0.01)^2$

$F=0.086\ N$

We need to calculate the acceleration

Using formula of force

$F=ma$

Put the value into the formula

$0.086=110\times a$

$a=\dfrac{0.086}{110}$

$a=0.000781\ m/s^2$

$a=7.81\times10^{-4}\ m/s^2$

(a). We need to calculate speed of the block

Using equation of motion

$v^2=u^2+2ad$

Put the value into the formula

$v=\sqrt{2\times7.81\times10^{-4}\times100}$

$v=0.395\ m/s$

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0

3 years ago

Why is it correct to say that the linear electron flow of the light reactions directly captures energy in nadph but only indirec

denis-greek [22]

Answer:

he electron is directly transferred to NADP+ to NADPH, but electron flow is used to generate a proton gradient for ATP synthesis. Electron is not directly transferred to ATP

NADP= Nicotin amide adenine dinucleotide phosphate

NADPH =Nicotinamide adenine dinucleotide phosphate is the reduced form of NADP

therefore electron is not directly transferred to ATP

5 0

3 years ago

What ocean depth would the volume of an aluminium sphere be reduced by 0.10%

yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0

3 years ago

A student tests the acidity of nitric acid (HNO3) by dissolving a sample of HNO3 in a solution of liquid methane. He then uses a

DerKrebs [107]

Sorry I didn't see this before...
Okay, I see two major problems with this student's experiment:

1) Nitric acid Won't Dissolve in Methane
Nitric acid is what's called a mineral acid. That means it is inorganic (it doesn't contain carbon) and dissolves in water.
Methane is an organic molecule (it contains carbon). It literally cannot dissolve nitric acid. Here's why:
For nitric acid (HNO3) to dissolve into a solvent, that solvent must be polar. It must have a charge to pull the positively charged Hydrogen off of the Oxygen. Methane has no charge, since its carbon and hydrogens have nearly perfect covalent bonds. Thus it cannot dissolve nitric acid. There will be no solution. That leads to the next problem:

2) He's Not actually Measuring a Solution
He's picking up the pH of the pure nitric acid. Since it didn't dissolve, what's left isn't a solution—it's like mixing oil and water. He has groups of methane and groups of nitric acid. Since methane is perfectly neutral (neither acid nor base), the electronic instrument is only picking up the extremely acidic nitric acid. There's no point to what he's doing.

Does that help?

8 0

3 years ago

Read 2 more answers

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals.

Bess [88]

time should you wait between pushes is 2.83 sec.

the question is incomplete, full statement is-

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?

<h3>What is Amplitude?</h3>

In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.

The length of swing, l = 2.1 m

The time between the pushes is nothing but the Time period

and is given by the formula,

$T = 2\pi ( \frac{l}{g} )^{\frac{1}{2} }$

= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)

= 2.83 sec

to learn more about Amplitude go to - brainly.com/question/3613222

#SPJ4

3 0

1 year ago