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3 years ago

Which one of the following frequencies of a wave in the air can be heard as an audible sound by the

Physics

2 answers:

balu736 [363]3 years ago

5 0

The correct answer would be the letter B because de , the human , can hear from 50 Hz to 20000 Hz of frecuency .

Flauer [41]3 years ago

4 0

B. 1,000 Hz Is the right answer.

You might be interested in

On a balanced seesaw, a boy three times as heavy as his partner sits

slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

$W_1 d_1 = W_2 d_2$

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

$W_1 = 3 W_2$

If we substitute this into the equation, we find:

$(3 W_2) d_1 = W_2 d_2$

and by simplifying:

$3 d_1 = d_2\\d_1 = \frac{1}{3}d_2$

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0

3 years ago

The sound wave passes from the sea-water into the air. State what happens to; the speed of the waves

dusya [7]

Answer. Explanation: Frequency of the sound decreases and the speed of sound becomes 346m/s from near about 1500 m/s.

6 0

2 years ago

Read 2 more answers

Emotions refer to the act or process of knowing. True or False

vaieri [72.5K]

Answer:

True

Explanation:

Because they are central to the human experience and are related to everything we do .

4 0

3 years ago

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

2 years ago

PLZ HELP I DON'T UNDERSTAND!! a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, t

algol13

Explanation:

The mass of a ball, m = 2 kg

It is traveling with a speed of 10 m/s

The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J$

The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.

Hence, the required kinetic and potential energy is 100 J.

6 0

2 years ago