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Bond [772]
3 years ago
9

Consider a cylindrical pipe with both ends open and fundamental resonance frequency 150 Hz. If this particular pipe has a length

of L, what the wavelengths of three lowest modes produced by this pipe? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. L,2L, 3L b. 2L, L, 2L/3 с. 2L/3, L, 2L d. L, L/2, L/3 e. 4L, 24, 4L/3 f. 4L, 4L/3, 4L/5 g. Neither of above.
Physics
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

answers, the correct one is b

Explanation:

For this problem we use the fact that since the ends are open at these points we have a maximum,

fundamental L =λ/2 λ = 2L

second harmonic L = λ λ = 2L / 2

third harmonic L = 3 λ/2 λ = 2L / 3

n harmonic lam = 2L / n

 with n an integer

therefore the length of the first three harmonics are: 2L, L, 2L / 3

when examining the different answers, the correct one is b

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A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af
Lady_Fox [76]

Answer:

<h3>2.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

4 0
4 years ago
What is meant by a colored shadow? If "shadow" means
Natasha_Volkova [10]

Explanation:

Red, green, and blue are therefore called additive primaries of light. ... When you block two lights, you see a shadow of the third color—for example, block the red and green lights and you get a blue shadow. If you block only one of the lights, you get a shadow whose color is a mixture of the other two.

First, your definition of a shadow is incorrect. A shadow is an area that receives less light than its surroundings because a specific source of light is blocked by whatever is "casting" the shadow. Your example of being outside reveals this. The sky and everything around you in the environment (unless you are surrounded by pitch black buildings) is sending more than enough light into your shadow, to reveal the pen to your eyes. The sky itself diffuses the sunlight everywhere, and the clouds reflect plenty of light when they are not directly in front of the Sun.

If you are indoors and have two light bulbs, you can throw two shadows at the same time, possibly of different darknesses, depending on the brightness of the light bulbs.

It can take a lot of work to get a room pitch black. One little hole or crack in some heavy window curtains can be enough to illuminate the room. There are very few perfectly dark shadows.

4 0
3 years ago
How would you set it up?
Firdavs [7]

Answer:

THE LAST OPTION

Explanation:

5 0
4 years ago
A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will th
Elena L [17]

Answer:

t=5.30s

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

y=v_{0y}t-\frac{gt^2}{2}

The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that v_{0y}=v_0sin\theta. Solving for t:

0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s

5 0
4 years ago
A rock is dropped from a 110-m-high cliff. How long does it take to fall (a) the first 55.0 m and (b) the second 55.0 m?
Genrish500 [490]

Answer:

a) t = 3.35[s]; b) t = 1.386[s]

Explanation:

We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].

We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.

And using kinematics equations:

v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]

Now we can calculate the time:

v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]

Now we can calculate the second time, but using as a initial velocity 32.85[m/s].

The final velocity will be:

v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]

Now we can calculate the second time:

t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]

Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.

7 0
3 years ago
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