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alex41 [277]
3 years ago
13

Compare and contrast infrasonic and ultrasonic vibrations

Physics
1 answer:
Olegator [25]3 years ago
8 0

Answer:

Explanation:

Ultrasonic waves are acoustic waves that are so high in frequency that humans can't hear them; however, infrasonic waves are sound waves that are lower in frequency than what humans can hear. A subsonic wave is a wave that is traveling slower than the speed of sound and a supersonic waves travels faster

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What is the magnite of the net displacement of the mouse?
agasfer [191]

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The  values is  s =  7.49 \  m

Explanation:

From the question we are told that

   The  distance it travels eastward is  x =  5.0 \ m

   The distance it travel towards the southeast  is l  =  3.0\  m

   The distance it travel towards the south is  z =  1 \  m

 

Let x-axis  be east

      y-axis  south

       z-axis into the ground

The angle made between east and south is  \theta  =  45^o

The displacement toward x-axis is

       x =  5 +  3cos(45)

       x =  7.12

 The  displacement toward the y-axis is  

     y  =  3 *  sin (45)

      y =  2.123

Now the overall displacement of the rat is mathematically evaluated as

        s =  \sqrt{7.12^2 +  2.12^2 +  1^2}

      s =  7.49 \  m

     

   

3 0
3 years ago
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?
Bas_tet [7]

Answer:

God is omnipresent.

Explanation:

This means God is everywhere and He works where ever we are in the world

3 0
3 years ago
Enumerate the function of each parts of alimentary canal in order from ingestion to egestion ​. ?
SOVA2 [1]

Answer:

Human digestive system involves gastrointestinal tract and other components such as liver, intestines, glands, mouth, stomach, gallbladder. The Human Digestive System Process has six primary tasks: ingestion, motility, secretion, digestion, absorption, excretion.

Explanation:

Hope it helps

3 0
2 years ago
A book is pushed across the table by a student. The student stops pushing on the book and the book slows down and stops. The for
VashaNatasha [74]
A. inertia 

Because it has to do with the motion of something, especially if it changes its pace. In this example, the book's motion, when sliding on the table, decreased because of less force being given off from the student. 
3 0
3 years ago
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