All categories

2 years ago

Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

Physics

1 answer:

Mars2501 [29]2 years ago

7 0

Answer:

hello your question has some missing parts below is the complete question

and the missing diagram

The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

answer : 1316.2 hertz

Explanation:

The frequency that produce the maximum sound intensity can be calculated using the relation below

dsin ∅ = n A

where A = dsin ∅ / n when n = 1 . d = 0.800

A = 0.800 * ( 1 / 3.162 )

A = 0.253 m

speed of sound = 333 m/s

frequency = speed / A

= 333 / 0.253 = 1316.2 hertz

You might be interested in

A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can

gavmur [86]

Answer:

d. at the same velocity

Explanation:

I will assume the car is also travelling westward because it was stated that the helicopter was moving above the car. In that case, it depends where the observer is. If the observer is in the car, the helicopter would look like it is standing still ( because both objects have the same velocity). If the observer is on the side of the road, both objects would be travelling at the same velocity. Also recall that, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes but speed is the rate at which object covers distance and it is not direction wise. Hence velocity is the best option.

5 0

2 years ago

An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in

vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0

2 years ago

If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?

Dmitrij [34]

E. Galaxy Cluster

Explanation:

A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.

A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.

A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.

Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.

Keywords: galaxies, parsec, megaparsec, galaxy cluster

Learn more about galaxy clusters and astronomical units from:

https://brainly.in/question/4624292

brainly.com/question/14214806

brainly.com/question/13315988

#learnwithBrainly

4 0

2 years ago

A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m

tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/ (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0

2 years ago

A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T

irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, $W_{FH} = 972J$

Force exerted, F' = 310 N

Angle, $\theta = 23^{\circ}$

Now,

The component of force acting horizontally is F'cos $\theta$

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

$W_{FH} = \vec{F'}.\vec{d}$

$972 = 310\times dcos23^{\circ}$

d = 3.406 m = 3.4 m

3 0

2 years ago