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2 years ago

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The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

s, how many resonances are between the two given ones, and if the pipe is closed at one end, how many resonances are between the two given ones?

1 answer:

Xelga [282]2 years ago

8 0

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{4}{9.81}$

μ = 0.408

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2 years ago

For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

$85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}$

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity ( $v_t$ ), in order to obtain the angular velocity of the wheel:

$v_t=w*R\\w=\frac{v_t}{R}$

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

$w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}$

And now, recalling that $2\pi$ radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by $2\pi$ :

rotational frequency = $\frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}$

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2 years ago

If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?

barxatty [35]

Answer:

$2.5\cdot 10^{-4}$

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

$S=\frac{\Delta L}{L_0}$

where

$\Delta L$ is the change in length of the object

$L_0$ is the original length of the object

In this problem, we have $L_0 = 4 m$ and $\Delta L=1 mm=0.001 m$ , therefore the strain is

$S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}$

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The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

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Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

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Ronch [10]

Answer

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