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Lady bird [3.3K]

2 years ago

14

Two loudspeakers in a plane, 5.0m apart, are playing the same frequency. If you stand 14.0m in front of the plane of the speaker

s, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 14.0m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.-What is the frequency of the sound? Assume a sound speed of 340 m/s.

1 answer:

Gre4nikov [31]2 years ago

3 0

Answer:

f= 196.3 Hz

Explanation:

The path difference between two waves given as

$\Delta = \sqrt{14^2+5^2}\ -14$

Δ = 0.866 m

We know that to hear minimum intensity the path difference given as

Δ = λ/2

λ = 2Δ

We know that

C= f λ

Now by pitting the values

340 = 2 x 0.866 x f

f= 196.3 Hz

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A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

(2) Angular frequency of the wave is given by :

$\omega=2\pi\nu$

$\omega=2\pi\times 4921.56\ Hz$

$\omega=30923.07\ rad/s$

(3) The period of oscillation is given by T as :

$T=\dfrac{1}{\nu}$

$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

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2 years ago

A jet engine gets its thrust by taking in air, heating and compressing it, and

nadya68 [22]

Answer:

$F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw$

Explanation:

Thrust is known as a reaction force which appears when a system expels or accelerates mass in one specific direction. If we know the acceleration and the mass of the air expelled by the jet engine, we can compute the thrust .

The acceleration is calculated by using the dynamics formula

$\displaystyle a=\frac{v_f-v_o}{t}$

The values are

$v_f=500\ m/s,\ v_o=100\ m/s,\ t=1\ sec$

$\displaystyle a=\frac{500-100}{1}=400\ m/s^2$

The thrust is

$F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw$

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2 years ago

Select the areas that would receive snowfall because of the lake effect.

Studentka2010 [4]

Answer:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The places that would would get snowfall because of the lake effect are Grand Marais, Two Harbors, and Duluth. The reason for this is that these three places are located right on the shores of the Lake Superior. This lake is one of the biggest lakes in the world. It has enormous amount of water in it, having big impact on the regional climate because of that. The water from this lake creates a lot of humidity in the air, and there's a lot of evaporation as well, both causing the formation of clouds, and when it is cold enough, instead of precipitation, the region gets large amounts of snowfall.

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2 years ago

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2

Tatiana [17]

You can reason it out like this:

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

   = 3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

   Distance = (1/2) · (acceleration) · (time)²

   = (1/2) · (8 m/s²)    · (30 sec)²

   = (4 m/s²) · (900 sec²)

   = 3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s². That's about 0.82 G !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.

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2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago