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3 years ago

15

Two organ pipes, both open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2 cm. Find the fre

quency of the beat they produce when playing together in their fundamental frequencies.
A) 0.9 Hz
B) 3.4 Hz
C) 2.4 Hz
D) 1.3 Hz

1 answer:

zloy xaker [14]3 years ago

4 0

Answer: 1.3 Hz

Explanation:

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List at least 3 ways for you to represent your data graphically

yawa3891 [41]

1. circle graph
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1 year ago

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Considera que en tu casa tienes un televisor de 110 W, si pasa encendido 4 horas diarias, Cuál será la energía consumida durante

rjkz [21]

Answer:

E = 13.2 kWh

, Cost = $ 10.8

Explanation:

We can look for the consumed energy from the expression of the power

P = W / t

The work is equal to the variation of the kinetic energy, for which

P = E / t

E = P t

look for the energy consumed in one day and multiply by the days of the month in the month

E = 110 4 30

E = 13200 W h

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3 years ago

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A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

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3 years ago

The spring constant for the spring in above Table is 20 N/m.

satela [25.4K]

The answer is 0.025J.

W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J

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3 years ago

an airplane traveling 245 m/s east expericences turbulence, so the pilot slows down to 230 m/s. it takes the pilot 7 seconds to

lana66690 [7]

Answer:

a=v-u/t

a=245-230/7

a=2

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2 years ago