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Lapatulllka [165]
3 years ago
7

For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. For what val

ue of the fundamental frequency of audible sound will these speakers produce constructive interference at the listener? The speed of sound is 344 m/sec. (1 point)
Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

We want a sound wave with a wavelength of 0.52 meters or a natural fraction thereof. We'll work in MKS.

w = 0.52/n

That's length.  We have speed 344 meters/second so w corresponds to a frequency of

f = 344 / w = n (344/.52)

f = 661.5 n  Hertz

I don't really agree with how they're saying it, but all the fundamental talk is probably trying to tell us n=1,

Answer: 661.5 Hertz

Any multiple of that will also produce constructive interference; we can go to about n=30 before we're out of the audio range.

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A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

F_n = 306 N

Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15 \times 9.81}

v = 12.1 m/s

So this speed is independent of the mass of the rider

5 0
4 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
A nonzero net force acts on a particle and does work. Which one of the following statements is true?
arlik [135]

Answer:

option (D) is correct.

Explanation:

According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.

the kinetic energy of a body is directly proportional to the square of the speed of the body.

As the kinetic energy change, the speed of the body also change.

Option (D) is correct.

5 0
3 years ago
What is the benefit of developing the atomic mass unit as a standard unit of mass?
Alexeev081 [22]

Answer:

The atomic mass unit is 1/12 of an atom of carbon 12, and is a very small amount to represent in kilograms:

1m_{u}=1.66054x10^{-27}kg

m_{u} is atomic mass unit.

This is why the benefits of the atomic mass unit is that it makes the representation of atomic masses easier in terms of the simplicity of the numbers that are used to represent the masses. Also using the atomic mass unit it is easier to compare the masses of different atoms, These numbers would be very small and would require negative powers of 10 to represent them, so it is more convenient to use the atomic mass unit.

3 0
3 years ago
What is a gravitational field and how its strength be measured
Yakvenalex [24]
A gravitational field is the field generated by a massive body, that extends into the entire space. Every object with mass m experiences a force F when immersed in a gravitational field. The intensity of the force is equal to
F= \frac{GM}{r^2}  m
where G=6.67 \cdot 10^{-11} m^3 Kg^{-1} s^{-2} is the gravitational  constant, M is the mass of the source of the field (e.g. the mass of a planet), and r is the distance between the object and the source of the field. The force is always attractive. 

A possible way to measure the intensity of a gravitational field is by measuring the acceleration a of the object immersed in this field. In fact, for Newton's second law we have:
F=ma
but since 
F= \frac{GM}{r^2} m
we can write
a =  \frac{GM}{r^2}
Therefore, by measuring the acceleration of the object, we also measure the intensity of the field.

5 0
3 years ago
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