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dimaraw [331]
2 years ago
9

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?​

Physics
1 answer:
swat322 years ago
4 0

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s

So, the speed of the wave is equal to 8.4 m/s.

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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0
3 years ago
Helpppppppp<br> ill give brainliest
Nina [5.8K]

Answer:

I believe the answer is

B. \: 3 \: red, \: 8 \: yellow, \: 1 \: blue

Explanation:

I may be incorrect because I used to do this a long time ago but I believe I am correct

HOPE THIS HELPS!

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4 0
2 years ago
A 100 kg marble slab falls off a skyscraper and falls 200 m to the ground without hitting anyone. Its fall stops within millisec
Radda [10]

Answer:

Δ T = 2.28°C

Explanation:

given,

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C_marble = 860 J/(kg °C)

using conservation of energy

Potential energy = heat energy

  m g h = m C_{marble}\Delta T

  g h =C_{marble}\Delta T

  \Delta T= \dfrac{g h}{C_{marble}}

  \Delta T= \dfrac{9.8 \times 200}{860}

        Δ T = 2.28°C

7 0
3 years ago
single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

4 0
3 years ago
A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us
Nataliya [291]

Answer:

T_C=118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= \frac{T_{H} }{T_{H}-T_{C} }

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP=\frac{Q_H}{W}

in the question COP= \frac{60}{100} \times COP_{max}

⇒\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}

heat is added directly to be as efficient as via heat pump

Q_H= W

and T_H= 24° C= 297 K

1=\frac{60}{100}\times \frac{297}{297-T_C}

on calculating the above equation we get

T_C=118.8 K

the outdoor temperature for efficient addition of heat to interior of home

T_C=118.8 K= 154.2°C

6 0
3 years ago
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