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Rina8888 [55]
3 years ago
5

Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A column of air has a hei

ght equal to 1⁄8 of the wavelength of the sound waves produced by a tuning fork vibrating over the column of air in an open pipe that's partially immersed in water. B. A person is holding a tuning fork on one end of a football field and another person is holding a separate tuning fork on the opposite end of the field. C. A pipe's length is equal to 1⁄2 of the wavelength of the sound waves produced by a tuning fork vibrating over one end of the pipe that's open to the air at both ends. D. A vibrating tuning fork is struck and begins to vibrate as the object used to strike it is placed away from the tuning fork.
Physics
1 answer:
ollegr [7]3 years ago
8 0
The correct option is D.
Resonance is a phenomenon in which a vibrating system or an external force induces another system to vibrate at greater amplitude at particular frequencies. 
When a tuning fork is struck with another object, the tuning fork will start to vibrate as a result of the force that is exerted by the object used to struck it. This type of vibration, without direct excitation by a man is an example of resonance.
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An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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