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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav

uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz

4 0

3 years ago

• Describe the importance of a commonly used measuring system. Be prepared to give two examples of why having a common system of

IrinaVladis [17]

Metric system. important because it saves us time due to the fact that we do not have to convert our units every time.

3 0

3 years ago

An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?

Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0

3 years ago

A constant current I circulates counterclockwise around a wire loop in the shape of a triangle, lying in x-y plane, with vertice

Blizzard [7]

Answer:

Considering a triangle like the one of the figures, you can obtain the total magnetic force applied on it like the addition of the forces applied to each of the 3 sides.

Been the magnetic force formula:

$F=\int\limits^a_b {I} \, \overline {dx} \times \overline {B}$

For each segment:

Segment 1 (from [0,0,0] to [a,0,0]):

$F=\int\limits^a_0 {I} \, \overline {x}dx \times B \overline {z}=-IBa\overline {y}$

Segment 2 (from [a,0,0] to [0,a,0]):

$F=\int\limits^{(0,a)}_{(a,0)} {I} \, \frac{\sqrt{2} }{2} (\overline {x}+\overline {y})dx \times B \overline {z}=IB2a(\overline {x}+\overline {y})$

Segment 3 (From [0,a,0] to [0,0,0]):

$F=\int\limits^0_a {I} \, \overline {y}dx \times B \overline {z}=-IBa\overline {x}$

Total force on the wire loop:

$F_{T} =F_{1} +F_{2} +F_{3} =-IBa\overline {y}+IB2a(\overline {x}+\overline {y})-IBa\overline {x}=0N$

5 0

3 years ago

In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mas

Aneli [31]

Answer:

a. ΔP/Δt = 42.6 N

b. F = 42.6 N

c. P = 142042.4 Pa = 1.42 KPa

Explanation:

a.

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

Vf = Final Speed = 1110 m/s

Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

ΔP/Δt = 42.6 N

b.

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

F = 42.6 N

c.

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

P = 142042.4 Pa = 1.42 KPa

4 0

3 years ago