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3 years ago

To what does the amplitude of a sound refer to?

Physics

1 answer:

Alex777 [14]3 years ago

7 0

Answer:

The number of molecules displaced by a vibration creates the amplitude of a sound. The strength or level of sound pressure. The number of molecules in the sound wave in b is greater than the number in the sound wave in a, therefore the amplitude of the sound wave in b is greater.

Explanation:

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What is the frequency of a wave

MAVERICK [17]

The frequency of a wave is the number of waves that passes through a point in a certain time. The less waves that pass in a period of time the lower the frequency of the wave. The more waves that pass in a period of time the higher the frequency of the wave. When measuring wave length the time period used is usually one second.

6 0

3 years ago

Read 2 more answers

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

3 0

3 years ago

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

Montano1993 [528]

Answer:

vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

W = ΔK (1)

for this part the mass is

M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

N -W = 0

N = W = Mg

friction forces have the expression

fr =μ N

fr = μ M g

we substitute in 1

-μ M g x = 0 - ½ M v²

v² = 2 μ g x

let's calculate

v² = 2 0.750 9.8 6.00

v = ra 88.5

v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

p_f = (mₐ + m_b) v

p₀ = p_f

+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

660 vₐ₀ = (660 +490) 9.39 + 490 17.778

vₐ₀ = 19509.72 / 660

vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0

3 years ago

You went 100 miles north and 14

skad [1K]

Answer:

82 degrees

Explanation:

consider your staying point to be the center of a circle. this center has the coordinates (0, 0).

the radius of the circle is the distance you walked East (14 miles).

I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.

then the 100 miles North is tan(displacement angle)×14.

as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.

so, again.

100 = 14×tan(angle)

tan(angle) = 100/14 = 50/7 = 7.142857...

the displacement angle is then 82 degrees.

5 0

3 years ago

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored, how high (in m) would

motikmotik

Then the maximum height of the catapult will be 69.78 meters.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion. The equation of motion is given below.

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored.

a = - 9.81 m/s²

u = 37 m/s

v = 0

s = h

Then the maximum height of the catapult will be given by the third equation.

v² = u² + 2as

0² = 37² - 2 × 9.81 × h

h = 37² / (2 × 9.81)

h = 69.78 meters

Then the maximum height of the catapult will be 69.78 meters.

More about the kinematics link is given below.

brainly.com/question/7590442

#SPJ1

4 0

1 year ago