Ask question

Ask question

All categories

2 years ago

10

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p

erson stands 4.1 m away from one speaker and 4.8 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 343 m/s.

1 answer:

serg [7]2 years ago

7 0

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m

For destructive interference

$d = \dfrac{n\lambda}{2}$

where, n = 1, 3,5..

we know, λ = v/f

$d = \dfrac{n v}{2f}$

v is the speed of the sound = 343 m/s

f is the frequency

$f = \dfrac{n v}{2d}$

for n = 1

$f = \dfrac{343}{2\times 0.7}$

f = 245 Hz

for n = 3

$f = \dfrac{3\times 343}{2\times 0.7}$

f = 735 Hz

Hence,the second lowest frequency of the destructive interference is 735 Hz.

You might be interested in

A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this

noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

The displacement of the object is d = 3.00 m, while $\theta=180^{\circ}$ , because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

$W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J$

And the work done is negative, because it is done against the motion of the object.

6 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is

Digiron [165]

All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is in a state known as constant velocity. The bullet is moving at a constant value of velocity. Acceleration is the rate of velocity so having zero acceleration would mean that there is no change in velocity per unit of time.<span />

8 0

2 years ago

Read 2 more answers

Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha

schepotkina [342]

The radius of the circular path is 1.5 m.

The circumference is then
$1.5\ m*2\pi=3\pi\ m$

The ball moves 3π m every 2.2 s, so the speed is
$\frac{3\pi\ m}{2.2\ s}\approx 4.3\ m/s$

9 0

2 years ago

Read 2 more answers

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago