1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Igoryamba
3 years ago
5

At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s

m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
Physics
1 answer:
Leto [7]3 years ago
4 0

Answer:

Angle = 0.2520 radians

Explanation:

Complete question:

Sound with frequency 1220Hz leaves a room through a doorway with a width of 1.13m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.

Given Data:

Speed of sound =v= 344 m/sec    ;

Width of doorway =d= 1.13m          ;

Frequency of sound =f= 1220 Hz  ;

Solution:

As we know that

Wvelength = w = v/f = 344/1220 = 0.281967m

Now we also know that

w = dsin(A)      where A is the angle

A = arcsin(w/d) =14.44° = 14.44*(3.14/180) = 0.2520 radians

At the angle of 0.252 radians relative to the centreline perpendicular to the doorway a person outside the room will hear no sound under given conditions.

You might be interested in
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
4 years ago
What causes tendonitis?
Irina18 [472]

overuse of a muscle Answer:

Explanation:

6 0
3 years ago
Read 2 more answers
HELP IM TIMED. ILL MARK BRAINLIEST PLSSSSSS.
Alex17521 [72]

Answer:

According to Newton's Second Law of Motion :    

Where,

F = Force Applied

m = Mass of the object

a = Acceleration

Now, we will use this law to solve this question.

Given :

Acceleration or a = 15.3 m/s²

Force = 44 N

Mass = ?

Substitute, the given values in the formula.

F = ma

⇒ m = F/a

   m = 44/15.3

<u>m = 2.9 kg</u>

5 0
3 years ago
Evaluate tan45/tan45
SIZIF [17.4K]

I'm not sure I completely understand the expression you want evaluated.

It looks like a fraction with the same exact thing in both the numerator and the denominator. A fraction like that always boils down to ' 1 '.

5 0
3 years ago
Read 2 more answers
Explain what causes acid rain to form and its negative effects on living and nonliving things.
OLga [1]
What causes it to form: <span>Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide and nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain. Human activities, power plants, and exhaust from vehicles all are causes.</span>
Affects on living things-
Acid rain can kill trees
It can affect plant growth patterns
Toxins can kill animals
Nonliving things- 
It can decay building materials and paints
it breaks down rocks

Hope this helped! Mark brainliest?
6 0
3 years ago
Other questions:
  • What formula relstes work n power
    11·1 answer
  • A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that th
    9·1 answer
  • The maximum pressure variations the human ear can withstand are about how much above and below atmospheric pressure?
    11·1 answer
  • a Stone is dropped into a deep well and is heard to hit the water 3.41s after being dropped. determine the depth of the well
    15·1 answer
  • A man throws a football straight into the air. As it rises, it slows down. Which
    8·1 answer
  • Why does physics involve math?
    14·1 answer
  • Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the waterÂs surface to knoc
    9·2 answers
  • A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
    11·1 answer
  • Which statement describes the path of electrons from a battery to a circuit?
    8·2 answers
  • so no body wants to help me:((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!