Ask question

Ask question

All categories

2 years ago

4

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

ng 3.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible

1 answer:

kherson [118]2 years ago

8 0

Answer:

1089.74 Hz

Explanation:

Using Pythagoras theorem, we can find the distance from the point to the second speaker.

Thus;

d2 = √(3² + 2.3²)

d2 = √(9 + 5.29)

d2 = √14.29

d2 = 3.78 m

Then, the path distance which is the extra distance travelled would be;

Δd = d2 - d1

Δd = 3.78 - 3

Δd = 0.78 m

Now, the destructive interference condition is given by the formula;

Δd = (m + ½)λ

λ is the wavelength

m is a non - negative integer.

In this case, m = 2

Thus;

0.78 = (2 + ½)λ

λ = 0.78/(2½)

λ = 0.312 m

Now the formula for frequency of a wave is given by;

f = v/λ

Where v is speed of sound.

Thus;

f = 340/0.312

f = 1089.74 Hz

You might be interested in

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha

Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0

2 years ago

A 0.242 g sample of potassium is heated in oxygen. The result is 0.292 g of a crystalline compound. What is the formula of this

masha68 [24]

Answer:

Hello there Dude answer is B :D hope it helped mark me brainliest.

8 0

2 years ago

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

GREYUIT [131]

Answer:

The flux through the surface of the cube is $2.314\ Nm^{2}/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0\times 10^{- 2}\ m$

Volume Charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$

Now,

To calculate the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$ (1)

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$ (2)

Volume of cube, $V = a^{3}$

Thus

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$

Thus from eqn (2), the total charge is given by:

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$

5 0

2 years ago

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago

Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o

Bond [772]

To answer the problem we would be using this formula which isv = sqrt(T/(m/L))
v = sqrt(100 N / [(0.100 kg)/(1.0 m)])
v = 31.62 m/s
v = fλ
31.62 m/s = (95 Hz)(λ)
λ = 0.333 m
For every wavelength along a string there will be 2 antinodes.
1.0 m / 0.333 m = 3
3 * 2 = 6 antinodes
6 + 1 = 7 nodes

4 0

2 years ago