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Pie
3 years ago
5

Suppose these waves represent the sound of a siren on a passing ambulance. Which wave represents the sound of the siren after it

has passed you? Explain your answer.
The wave A has a higher frequency and wave B has a lower frequency.

I REALLY need help. Please.
Physics
2 answers:
sergeinik [125]3 years ago
4 0
The answer is wave B has a lower frequency.  Suppose these waves represent the sound of a siren on a passing ambulance. Wave B (lower frequency) represents the sound of the siren after it has passed you.  The ambulance moving towards the observer produces a higher than normal frequency, and the ambulance moving away produces a lower than normal frequency.
sergey [27]3 years ago
4 0

Answer:

The correct answer is "wave B has a lower frequency."

Explanation:

The Doppler effect is defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave movement when the sender and the receiver, or observer, move relative to each other.

This is the effect that occurs in this case. And it is explained as follows: When the ambulance approaches the observer, the wave "compresses", that is, the wavelength is short, the frequency high and, therefore, the tone of the perceived sound will be sharp. On the other hand, when the ambulance moves away from the observer, the wave "decompresses", that is, the wavelength is long and the frequency low.

Then the correct answer is "wave B has a lower frequency."

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A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.
Ostrovityanka [42]

Answer:

a) 8 seconds if you are using earth's gravity.

b) 48m if the velocity does not change

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Explanation:

3 0
3 years ago
Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh
Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

N=N_{0}e^{-\lambda t}

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

514=1293e^{-0.0314t}

2.5155=e^{0.0314t}

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0
3 years ago
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