Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 641Hz . You are standing between the speakers, along the line connecting them and are at a point of constructive interference.How far must you walk toward speaker B to move to reach the first point of destructive interference? Take the speed of sound to be 344 .
You must walk approximately 0.2685 m, or 26.85 cm, towards speaker B to encounter the first point of destructive interference. This calculation is arrived at by determining the half-wavelength of the sound wave.
Interference occurs when two sound waves from the same source meet. When they constructively interfere, their amplitudes add together creating a louder sound, while when they destructively interfere, they cancel each other out creating a point of silence. Since you are initially in a position of constructive interference, you need to move towards speaker B at a distance that would change the path length difference to be equivalent to a half wavelength.
To find this distance, we first need to find the wavelength from the frequency. The formula for this is:
Wavelength = Speed of sound / Frequency
Given the speed of sound is 344 m/s and the frequency is 641 Hz, we find the wavelength to be roughly 0.537 m. A half wavelength, which characterizes the distance needed for destructive interference from total constructive interference, would then be 0.2685 m.
You must walk approximately 0.2685 m, or 26.85 cm, towards speaker B to encounter the first point of destructive interference.
To find the distance at which the first point of destructive interference occurs, divide the wavelength by 2. In this case, the distance is approximately 0.268 meters or 26.8 centimeters. Therefore, you would need to walk about 26.8 centimeters toward speaker B to reach the first point of destructive interference.
To determine the distance at which the first point of destructive interference occurs, we need to understand the concept of interference and the conditions for constructive and destructive interference. Constructive interference occurs when the waves from both speakers are in phase and add up to create a larger amplitude. Destructive interference occurs when the waves from both speakers are out of phase and cancel each other out, resulting in a smaller amplitude. In this case, since the speakers are emitting waves in phase, the distance at which destructive interference occurs is equal to half the wavelength of the waves.
The wavelength of a wave can be calculated using the formula: Wavelength = Speed of sound / Frequency
In this case, the frequency is given as 641 Hz and the speed of sound is given as 344 m/s. Plugging in these values into the formula, we get: Wavelength = 344 m/s / 641 Hz
Solving this, we find that the wavelength is approximately 0.536 meters. To find the distance to the first point of destructive interference, we divide the wavelength by 2: Distance to first point of destructive interference = Wavelength / 2
Plugging in the calculated wavelength, we get: Distance to first point of destructive interference = 0.536 meters / 2
Simplifying, we find that the distance is approximately 0.268 meters or 26.8 centimeters. Therefore, you would need to walk about 26.8 centimeters toward speaker B to reach the first point of destructive interference.
Learn more about distance of destructive interference here:
Assume: The bullet penetrates into the block and stops due to its friction with the block.
The compound system of the block plus the
bullet rises to a height of 0.13 m along a
circular arc with a 0.23 m radius.
Assume: The entire track is frictionless.
A bullet with a m1 = 30 g mass is fired
horizontally into a block of wood with m2 =
4.2 kg mass.
The acceleration of gravity is 9.8 m/s2 .
Calculate the total energy of the composite
system at any time after the collision.
Answer in units of J.
Taking the same parameter values as those in
Part 1, determine the initial velocity of the
Answer in units of m/s.
To solve this problem we will start considering the total energy of the system, which is given by gravitational potential energy of the total of the masses. So after the collision the system will have an energy equivalent to,
= mass of bullet
= Mass of Block of wood
The ascended height is 0.13m, so then we will have to
PART B) At the same time the speed can be calculated through the concept provided by the conservation of momentum.
Since the mass at the end of the impact becomes only one in the system, and the mass of the block has no initial velocity, the equation can be written as
The final velocity can be calculated through the expression of kinetic energy, so
Using this value at the first equation we have that,
If the velocity of a pitched ball has a magnitude of 41.0 m/s and the batted ball's velocity is 50.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
The change in momentum is (91 m/s) multiplied by the mass of the ball (which you neglected to mention). That's exactly the impulse delivered by the bat.
If a small child swallowed a safety pin, whywould an X-ray photograph clearly show the location of the pin?
it is in the body system
it would show clearly because it is a metal piece in the body.
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B
r₁/r₂ = 1/2 = 0.5
The resistance of a wire is given by the following formula:
R = ρL/A
R = Resistance of wire
ρ = resistivity of the material of wire
L = Length of wire
A = Cross-sectional area of wire = πr²
r = radius of wire
R = ρL/πr²
FOR WIRE A:
R₁ = ρ₁L₁/πr₁² -------- equation 1
FOR WIRE B:
R₂ = ρ₂L₂/πr₂² -------- equation 2
It is given that resistance of wire A is four times greater than the resistance of wire B.
R₁ = 4 R₂
using values from equation 1 and equation 2:
ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²
since, the material and length of both wires are same.
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L
ρL/πr₁² = 4ρL/πr₂²
1/r₁² = 4/r₂²
r₁²/r₂² = 1/4
taking square root on both sides:
r₁/r₂ = 1/2 = 0.5
The ratio of the radius of wire A to the radius of wire B is 1/2.
The resistance of a wire is given by the formula R = ρl/A, where R is resistance, ρ is resistivity, l is length, and A is the cross-sectional area of the wire. When the wire has a circular cross-section, the area can be calculated by the formula A = πr². The resistance of the wire then becomes: R = ρl/(πr²). If the resistance of wire A is four times that of wire B, we can set up the equation 4RB = RA. Substituting the expression for resistance, we get 4(ρl/(πrB²)) = ρl/(πrA²). Simplifying, we find that the ratio of the radius of wire A to the radius of wire B is one-half, or rA/rB = 1/2.
Learn more about Resistance and Radius Ratio here:
The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm
The center of mass of the bent wire is approximately 11.18 cm from the bend.
In order to find the center of mass of the bent wire, we need to divide it into two segments: the horizontal segment and the vertical segment. The length of each segment is half of the total length of the wire, which is 20 cm, so each segment is 10 cm long.
The center of mass of the horizontal segment is located exactly at its middle point, which is 5 cm from the corner. The center of mass of the vertical segment is also located at its middle point, which is 10 cm from the corner. Since the horizontal and vertical segments are orthogonal, the distance from the bend to the center of mass of the bent wire is the hypotenuse of a right triangle with legs of length 5 cm and 10 cm. Using the Pythagorean theorem, we can calculate the distance:
d = sqrt(5^2 + 10^2) = sqrt(25 + 100) = sqrt(125) = 11.18 cm
Therefore, the center of mass of the bent wire is approximately 11.18 cm from the bend.
wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
The electric field within a parallel plate capacitor with dielectric is given by:
is the surface charge density
k is the dielectric constant
is the vacuum permittivity
The area of the plates in this capacitor is
while the charge is
So the surface charge density is
The electric field is
So we can re-arrange eq.(1) to find k:
The surface charge density induced on each dielectric surface is given by
is the initial charge density
k = 7.18 is the dielectric constant
And by multiplying by the area, we find the charge induced on each surface: