Ask question

Ask question

All categories

4 years ago

5

Two train whistles have identical frequencies of 1.64 102 Hz. When one train is at rest in the station and the other is moving n

earby, a commuter standing on the station platform hears beats with a frequency of 4.00 beats/s when the whistles operate together. What are the two possible speeds that the moving train can have?

1 answer:

Amiraneli [1.4K]4 years ago

6 0

Answer:

Vs = 6.73 m/s or Vs = 16.3 m/s

Explanation:

frequency of the trains whistle (f) = 1.64 x 10^{2} Hz = 164 Hz

frequency of beats heard = 4 beats/s = 4 Hz

velocity of the stationary train (Vr) = 0

velocity of sound in air (V) = 343 m/s

velocity of the moving train (Vs) = ?

we can get the velocity of the moving train from the formula below

Fn = f x $\frac{V + Vr}{V - Vs}$ ...equation 1

where Fn = net frequency

case one - assuming the train is approaching the station Fn = 164 + 4 = 168 Hz

substituting the known values into equation 1

168 = 164 x $\frac{343 + 0}{343 - Vs}$

1.02 = $\frac{343 + 0}{343 - Vs}$

Vs = $343 - \frac{343 + 0}{1.02}$

Vs = 6.73 m/s

case two - assuming the train is leaving the station Fn = 164 - 4 = 160 Hz

substituting the known values into equation 1

168 = 160 x $\frac{343 + 0}{343 - Vs}$

1.05 = $\frac{343 + 0}{343 - Vs}$

Vs = $343 - \frac{343 + 0}{1.05}$

Vs = 16.3 m/s

You might be interested in

Are S waves transvers? (Need answer fast, hurry)

pychu [463]

Answer:

Yes they are transverse

Explanation:

7 0

3 years ago

Read 2 more answers

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial s

Brrunno [24]

(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

<h3>Conservation of mechanical energy</h3>

The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;

ΔK.E = ΔP.E

¹/₂m(v²- u²) = mg(hi - hf)

¹/₂(v²- u²) = g(0 - hf)

v² - u² = -2ghf

v² = u² - 2ghf

where;

v is the final velocity at upper level
u is the initial velocity
hf is final height
g is acceleration due to gravity

when u² = 2gh, then v² = 0,

when gravity reduces, u² > 2gh, and v² > 0

Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

<h3>Final speed</h3>

v² = u² - 2ghf

where;

u is the initial speed = 5 m/s
g is acceleration due to gravity and its less than 9.8 m/s²
v is final speed
hf is equal height

Since g on Epislon is less than 9.8 m/s² of Earth;

5² - 2ghf > 3 m/s

Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

5 0

3 years ago

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

$a=2.5 m/s^2$

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

$d=0+\frac{1}{2}(2.5)(35)^2=1531 m$

8 0

3 years ago

Which statement best describes the motion of air particles in a sound wave?

Elodia [21]

Answer:

The answer is D.

Explanation:

They vibrate parallel to the wave.

During the propagation of a sound wave in air, the vibrations of the particles are most accurately represented as longitudinal. Longitudinal waves are waves in which the motion of the individual particles of the medium occurs in a direction that is parallel to the direction of energy transmission.

8 0

2 years ago

Read 2 more answers