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3 years ago

14

if a source of sound waves is rabidly approaching a person the sound heard by the person appears to have?

2 answers:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

have a frequency. higher than the frequency actually produced by the source.

Explanation:

solong [7]3 years ago

7 0

Answer:

If a source of sound waves is rapidly approaching a person, the sound heard by the person appears to have - a pitch lower than the original pitch.

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Each element has an atomic number state which is meant by atomic number

netineya [11]

Answer: The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons. The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms.

Explanation:

3 0

2 years ago

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

bulgar [2K]

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

$P = 2\pi r$

Where,

r= radius

While centripetal acceleration is defined by

$a=\frac{v^2}{r}$

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

$\frac{P}{4} = \frac{2\pi r}{4}$

The same route you take is the distance traveled, that is

$x = \frac{P}{4}$

$x = \frac{2\pi r}{4}$

$1725 = \frac{2\pi r}{4}$

$r = 1098.17m$

PART B)

With the radius is possible calculate he centripetal acceleration,

$a = \frac{v^2}{r}$

$a = \frac{115^2}{1098.17}$

$a = 12.04m/s^2$

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of $12.04m/s^2$

3 0

3 years ago

What is the kinetic energy of a golf ball with a mass of 0.046 kg traveling at 15.5<br> m/s?

Serjik [45]

Answer:

<h2>5.53 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 0.046 \times {15.5}^{2} \\ = 0.023 \times 240.25 \\ = 5.52575$

We have the final answer as

<h3>5.53 J</h3>

Hope this helps you

3 0

3 years ago

A playground merry-go-round spins about its axis with negligible friction. A child moves from the center of the merry-go-round t

Juli2301 [7.4K]

Answer:

B

Explanation:

In this process on merry go round there is not any external torque so angular momentum will be conserve. Mass is always conserved.

4 0

2 years ago

An electron is to be accelerated from a velocity of 4.50×106 m/s to a velocity of 9.00×106 m/s. Through what potential differenc

s344n2d4d5 [400]

Answer:

-2.85 * 10^(-17) J

Explanation:

Parameters given:

Final velocity, v = 9 * 10^6 m/s

Initial velocity, u = 4.5 * 10^6 m/s

Using the conservation of energy formula, total energy is conserved:

K.Ein + PEin = KEf + PEf

K.Ef - K.Ein = P.Ein - P.Ef

=> -∆P.E = K.Ef - K.Ein

∆P.E = K.Ein - K.Ef

∆P.E = ½mu² - ½mv²

∆P.E = ½m[(4.5 * 10^6)² - (9 * 10^6)²]

∆P.E = ½ * 9.31 * 10^(-31) * (-61.25 * 10¹²)

∆P.E = -2.85 * 10^(-17) J

7 0

3 years ago