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3 years ago

8. What is the frequency heard by a person driving at 18 m/s toward a blowing factory whistle (600 Hz) if

Physics

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Answer:

631.7 Hz

Explanation:

The Doppler effect is a phenomenon that occurs when there is relative motion between an observer and a source of a wave. When this occurs, the is a change in the apparent frequency of the wave, as observed by the observer.

In particular, the new apparent frequency can be calculated as:

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the real frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if the observer is moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (negative if the source is moving towards the observer, positive if moving away)

In this problem:

v = 340.6 m/s is the speed of the sound wave

f = 600 Hz is the real frequency of the sound

$v_o = +18 m/s$ is the velocity of the person

$v_s=0$ is the velocity of the source

Substituting, we find:

$f'=\frac{340.6+18}{340.6}(600)=631.7 Hz$

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A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

When testing a home for radon, where is the most likely place for the highest detected level? A) the basement, due to ground see

german

Answer:

Explanation:

ive been buying loui i been shoppin

5 0

3 years ago

Read 2 more answers

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago

An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03

Mnenie [13.5K]

Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium $\alpha =23.6\times 10^{-6}/^{\circ}C$

We know that change in length is given by $\Delta L=L\alpha \Delta T$

So $0.03=35\times 23.6\times 10^{-6}\Delta T$

$\Delta T=36.32^{\circ}C$

So final temperature $=T_I+\Delta T=17+36.32=53.3196^{\circ}C$

5 0

3 years ago

As charges move in a closed loop, they

mamaluj [8]

As charges move in a closed loop, they gain as much energy as they lose.

<h3>What is principle of conservation of energy?</h3>

According to the principle of conservation of energy, in a closed or isolated system, the total energy of the system is always conserved.

The energy gained by the particles or charges in a closed system is equal to the energy lost by the charges.

Thus, we can conclude the following based on principles of conservation of energy;

As charges move in a closed loop, they gain as much energy as they lose.

Learn more about conservation of energy here: brainly.com/question/166559

8 0

2 years ago