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Art [367]
3 years ago
5

Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 4.0 mm

, then forward by 2.0 mm. Afterward, constructive interference is observed 1414, 1212, and 3434 the distance between the speakers along the line that joins them, and at no other positions along this line. Part A What is the maximum possible wavelength of the sound waves ?
Physics
1 answer:
loris [4]3 years ago
6 0

Answer:

4.4721m

Explanation:

#Use Pythagorean theorem to find the distance between the two speakers:

a^+b^2=c^2\\4^2+2^2=d^2\\\sqrt{20}\ mm=d

There are antinodes 1/4,1/2 and 3/4 of the distance between speakers.

The greatest antinode is 3/4-1/4=1/2

#Distance between consecutive antinodes is:

0.5\times \sqrt{20}=\lambda/2\\\\\lambda=4.4721m

Hence, the maximum possible wavelength of the sound waves is 4.4721m

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The quarterback throws a football at 25 m/s at a certain angle above the horizontal. If it took the ball 2.5 s to reach the top
otez555 [7]

Answer:

The ball was in the air for 5.0 seconds

Explanation:

When the ball is thrown above at a certain angle from the horizontal it follows a projectile motion.

The ball has a specific vertical and horizontal velocity.

The time take for the ball to reach at the top of its path is also the time taken by the vertical velocity vector to become zero. This is the first half of the projectile motion (ball is going up).

For the second half, the ball comes down and it accelerates with same gravitational acceleration from which it was decelerating in the first half. Hence the time to come down will be the same 2.5 seconds.

So the total time the ball was in the air is twice of 2.5, i.e,

5.0 seconds

3 0
4 years ago
A 0.20 kg stone is dropped (from rest) from a height of 40.0 m above the ground. How far above the ground is the stone when it's
Jlenok [28]
The farthest is 12


And the ground has to be 30
7 0
3 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
A force of 5N produces an acceleration of 8m/s2 on mass m1, and an acceleration of 24m/s2 on a mass m2. What acceleration would
kotykmax [81]

The acceleration that the same force will provide if both masses are tied together is; 6.0 m/s².

<h3>How to find the Acceleration?</h3>

We are given;

Force; F = 5 N

Acceleration of the first mass, a₁ = 8.0 m/s²

Acceleration of the second mass, a₂ = 24 m/s²

Formula for force is;

F = ma

Let us find both masses; m₁ and m₂.

m₁ = F/a₁

m₂ = F/a₂

Thus;

m₁ = 5/8 kg

m₂ = 5/24 kg

Total mass is; m = m₁ + m₂

m = 5/8 + 5/24

m = 15 + 5/24

m = 20/24 kg

Thus, acceleration if they are both tied together is;

a = F/m

a = 5/(20/24)

a = 6.0 m/s².

Read more about Acceleration at; brainly.com/question/605631

#SPJ1

4 0
2 years ago
A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
yulyashka [42]

Answer:

T=245.76N

Explanation:

We know that the frequency of the nth harmonic is given by f_n=nf, where f is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N

6 0
3 years ago
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