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Korvikt [17]
3 years ago
8

An experimental analysis of the natural oscillations in a particular structure shows the dominant frequencies of interest appear

at below 200 Hz. However, frequency information also exists at 350, 450, and 750 Hz. If the signal is sampled at 400 Hz, how will the information in the aliased frequencies appear in the sampled data

Physics
1 answer:
IgorLugansk [536]3 years ago
6 0

Answer:

to prevent the aliased frequency from  occurring at ±50Hz , WE MUST SAMPLE AT fs of min

⇒ 2 x 750

= 1500Hz

Explanation:

check attachment for explanations.

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3 years ago
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The correct choice is <em> A </em>.  Looking at the picture did the trick !

7 0
2 years ago
Can someone answer this for me? it’s due today and i need it done asap! i will give brainlist!!
Setler79 [48]

Answer:

1. The sound waves are longitudinal because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.

2. A pulse or a wave is introduced into a slinky when a person holds the first coil and gives it a back-and-forth motion. This creates a disturbance within the medium; this disturbance subsequently travels from coil to coil, transporting energy as it moves.  

Explanation:

5 0
2 years ago
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A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto
sesenic [268]
<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=l\times b\times h=5.2\times 2.4\times h=12.48h

Density of water=10^3kg/m^3

Weight of water displaced by barge=Buoyant force=-Weight of horse

Volume\;of\;water\times density\;of\;water\times g=410\times g

12.48h\times 1000=410

h=\frac{410}{12.48\times 1000}=0.03 m

1 m=100 cm

0.03 m=0.03\times 100=3cm

Hence, the depth of barge float=3 cm

<h2 />
4 0
3 years ago
Read 2 more answers
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