1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Harman [31]
4 years ago
15

What is the lowest level of sound that can be detected by human hearing? 0 Hz 10 Hz 20 Hz 100 Hz

Physics
2 answers:
maxonik [38]4 years ago
8 0
These choices are not "levels" of sound.  They're frequencies of sound.

The lower limit of audible frequency is usually considered to be 20 Hz.
But there are some people with ears that can hear 10 Hz, and others ...
particularly the elderly ... who can't hear anything below 100 Hz.

The question is a rather poor one.
Ad libitum [116K]4 years ago
8 0
For a person with normal hearing, when it comes to pitch the human hearing range starts low at about 20 Hz.
You might be interested in
What is power??????????
KatRina [158]
D.) The energy needed to create work.
8 0
3 years ago
Which statement is not true about light? 1 point It is a particle and a wave It has mechanical and transverse waves It has mecha
belka [17]

Answer:

Option C

Explanation:

Light is an electromagnetic wave as it has characteristics of both electrical and magnetic waves. It is not a mechanical wave. Water and sound wave are mechanical wave as they need some medium to travel while light do not need any medium to travel.

Hence, option C is correct

4 0
3 years ago
The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is
kirill [66]
Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
                    L = k/(d^2)
For Case 1,
                       L1 = k/(d1^2)
For Case 2,
                       L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2. 
Therefore, the loudness at your friend's position is 16 times that of yours. 
5 0
4 years ago
Read 2 more answers
A 0.25 kg ball bearing is released from rest at the surface of the water in the tank of water.
alina1380 [7]

Answer:

amount of work done = -1.77 J

Explanation:

given data

mass = 0.25 kg

speed = 5 m/s

height = 2m

to find out

amount of work done by the water on the ball bearing

solution

we get here potential energy when ball fall

potential energy = mgh

potential energy = 0.25 × 9.8 × 2

potential energy = 4.9 J

and that lose potential energy gain by kinetic energy that is express as

kinetic energy = 0.5 × m × v²

kinetic energy = 0.5 × 0.25 × 5²

kinetic energy = 3.13 J

so we get now work done that is

work done = kinetic energy - potential energy

work done = 3.13 J - 4.9 J

amount of work done = -1.77 J

6 0
3 years ago
An 8000- kg locomotive moving with a speed of 2.0 m/s on straight level tracks collides with a 15,000 kg stationary car. After t
Charra [1.4K]

Answer:

Total momentum before collision = 16 × 10³ kg.m/s

Total momentum after collision = 16,000 kg.m/s

Velocity after collision = 0.696 m/s

total Kinetic energy before collision = 16,000 J

Total Kinetic energy after collision = 5,600 J

Explanation:

This Kind of collision is called perfectly Inelastic Collision, because both objects stick together, and move in the same direction after collision.

Using the principle of conservation of momentum, which states that in a system, total momentum before collision is the same as total momentum after collision, the Net momentum of the system can be written as:

momentum before collision = momentum after collision

M₁V₁ + M₂V₂ = (M₁ + M₂)Vₓ

where:

M₁ = Mass of locomotive = 8,000 kg

V₁ = speed of locomotive = 2.0 m/s

M₂ = Mass of car = 15,000 kg

V₂ = speed of car = 0 m/s (stationary car)

Vₓ = final velocity after collision

The total momentum before collision = M₁V₁ + M₂V₂

(8,000 × 2) + (15,000 × 0) = 16,000 + 0 = 16,000 kg.m/s

∴ The total momentum before collision = 16,000 kg.m/s = 16 × 10³ kg.m/s

since Total momentum is conserved, momentum after collision = 16,000 kg.m/s

Finding velocity after collision

M₁V₁ + M₂V₂ = (M₁ + M₂)Vₓ

= (8,000 × 2) + (15,000 × 0) = (8,000 + 15,000) Vₓ

16000 + 0 = (23,000)Vₓ

23,000Vₓ = 16,000

∴ Vₓ = 16,000 ÷ 23,000 = 0.696 m/s

Velocity after collision = 0.696 m/s

Finally the total Kinetic energy before collision = \frac{1}{2} mv^{2}

= \frac{1}{2} * 8000 * 2^{2}

= 4000 × 4 = 16,000 J

∴ Finally the total Kinetic energy before collision = 16,000 J

Total Kinetic energy after collision = \frac{1}{2} mv^{2}

= \frac{1}{2} * (8000 + 15000) * (0.696)^{2}

\frac{1}{2} * 23000 * 0.4844\\

= 11,500 × 0.4844 = 5570.6 ≈ 5,600 Joules

5 0
3 years ago
Other questions:
  • Question 20
    9·1 answer
  • Which statement does Einstein's theory promote about space and time?
    6·2 answers
  • A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the
    9·1 answer
  • A 0.144kg baseball is pitched horizontally at 38.0 m/s. After the baseball is hit by the bat, it moves at the same speed, but in
    11·1 answer
  • Calculate the applied force of the washers on the car. First, convert the mass you recorded for one, two, three, and four washer
    8·2 answers
  • Gravitational potential energy is related to–A. what an object is made of. B.an object’s motion.C. an object’s position.D. an ob
    7·1 answer
  • The speed of a 1200 kg car increases from 5 to 29 m/s in 12 s.<br> What force accelerated the car?
    12·1 answer
  • If the speed of an object does not change the object Is traveling at _____speed
    8·1 answer
  • Please show your workmkmmkekdmnesnndndndndn
    9·1 answer
  • Help please and thank you
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!