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elena-14-01-66 [18.8K]

2 years ago

15

The __ is responsible for determining the frequency of vibration of the air column in the pool with in a wind instrumen . A. Typ

e of reed use . B. Effective length of the tube. C humidity of the air

2 answers:

BlackZzzverrR [31]2 years ago

5 0

The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument.

VladimirAG [237]2 years ago

3 0

Answer;

B. Effective length of the tube.

<h3><u>Explanation;</u></h3>

<u><em>Most wind instruments are of either the open-open tube or the open closed tube type. Open ends of the tubes makes contact with the outside, atmospheric air.</em></u>
At an open end, when the air is disturbed in the wind instruments, the air vibrates with a variety of frequencies, however the frequencies that correspond to the natural frequencies will persist.
<em><u>Effective Length of the tube is responsible for determining the frequency of vibration of the air column in the pool with in a wind instrument.</u></em>

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A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

3 years ago

A reactor operating at 1 MW is scrammed by instantaneous insertion of $5.00 of negative reactivity. Approximately how long does

Novosadov [1.4K]

Answer:

time is 3333.33 min or 55.55 hr

Explanation:

given data

reactor operating = 1 MW

negative reactivity = $5

power = 1 miliwatt

to find out

how long does it take

solution

we know here power coefficient that is

power coefficient = $\frac{10^{6} }{10^{6} }$

power coefficient = 1

so time required to reach power is

power = reactivity × time / power coefficient + reactor operating

1 × $10^{-3}$ = -5 t / 1 + 1 × $10^{6}$

5t = $10^{6}$ - $10^{-3}$

t = 199999.99 sec

so time is 3333.33 min or 55.55 hr

4 0

3 years ago

Why does sound propagate faster in solid bodies than in liquids and faster in liquids than in air?

storchak [24]

Answer:

see below

Explanation:

this is because particles in solids are packed very closely together, thus , the particles collide with each other frequently and thus transfer of energy is faster. however, particles in liquid are closely packed but not as close as in solid so the particles do not collide as frequently. thus, transfer of energy slower than in solid. furthermore, the particles in gas are spaced far apart from each other, thus the particles don't collide with each other frequently, thus transfer of energy is very slow in gas.

hope you get it,

please mark

4 0

3 years ago

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$

$\hat{ECD} = tan^{-1} 2 = 63.43^o$

This is also the angle ACB, so we can find the length AB:

$tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}$

$2 = \frac{AB}{234}$

$AB = 2*234 = 468 m$

So the height of the building is 468m

5 0

3 years ago