Ask question

Ask question

All categories

3 years ago

7

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

opposite to the first train and approaching it?

1 answer:

Sergio039 [100]3 years ago

7 0

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

Put the value into the formula

$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

You might be interested in

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

sattari [20]

Answer:

In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave.

6 0

2 years ago

W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively

MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

ΔU = ΔK

7 0

3 years ago

3. If a vector that is 1 cm long represents a

prisoha [69]

Answer:

A velocity of $20 \: km/h$

Explanation:

Since the magnitude of the vector is equal to the magnitude of velocity, velocity of the 2 cm vector represents a velocity of $10\times 2= 20 \: km/h$.

8 0

3 years ago

The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its

Scilla [17]

<h2>Power of cheetah is 5576.85 W = 7.48 hp</h2>

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44 ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

7 0

3 years ago

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

a)

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

e)

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

7 0

3 years ago