20 POINTS!

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

e-lub [12.9K]

Answer:

$f = 421.8 Hz$

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

$\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}$

so it will be

$\Delta L = 12.5 - 12.093$

$\Delta L = 0.4066$

now we know that

$\frac{\lambda}{2} = 0.4066$

$\lambda = 0.813$

now frequency of sound is given as

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.813}$

$f = 421.8 Hz$

4 0

3 years ago

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the

Vesna [10]

Answer:

The height of the cliff from which the ball was dropped from is 224.4m.

\overline{v}={\frac{\Delta x}{\Delta t}}

Given the data in the question;

Initial velocity of the ball;

Time taken by the ball to reach the ground;

Distance or Height of the cliff from which the ball was thrown from;

To get the height of the Cliff, we use the Second Equation of Motion:

Where s is the distance or height, is the initial velocity, t is the time and a is the acceleration. Since the ball was thrown down from a certain height (cliff), its is now under the influence of gravity. acceleration due to gravity;

Hence, the equation becomes

We substitute the given values into the equation

Therefore, the height of the cliff from which the ball was dropped from is 224.4m

Explanation:

3 0

2 years ago

Read 2 more answers

This is for physics, i need help fast!!

jarptica [38.1K]

Speed = distance / time
S= 40 000m / 5400s
S= 7.41m/s

7 0

3 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$