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Dominik [7]
3 years ago
9

What is the magnitude of the detected sound frequency shift from 170 Hz during the projectile flight described in the passage?

Physics
1 answer:
Serhud [2]3 years ago
5 0

Answer:

The magnitude of the frequency shift (that is the change in frequency) first falls to zero, then increases. This is so because as the projectile moves upward, the position of the sound source is constantly changing and as a result there would be a difference between the frequency of the sound emitted and the sound received by the microphone (that is the frequency shift). A point will be reached in the projectile's motion when the projectile will be momentarily at rest. (At the highest point reached in its flight h).

Explanation:

At this point, the frequency emitted by the source becomes equal to that received by the microphone. This can be seen from the equation 2 for Doppler effect contained in the attachment below.

This moment is short lives and the projectile soon starts a return journey downward under the influence of gravity. The velocity of the projectile is negative ( v < 0 ) and from the equation can be confirmed that the frequency received by the microphone increases as the projectile fall under gravity. This is so because as the projectile falls downward it velocity becomes more and more negative. And from the equivalent in the picture below, the frequency received by the microphone increases

Thank you very much and I hope this was helpful.

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What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the oth
Mandarinka [93]

Answer:

Explanation:

Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)

MA = Li/Le

Given;

Li = 45cm

Lo = 1.8cm

MA = 45/1.8

MA = 25

Hence the mechanical advantage is 25

Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.

MA = Load/Effort

Given

Load = 1250N

MA = 25

Effort = ?

Substitute

25 = 1250/Effort

Effort = 1250/25

Effort = 50N

Hence the minimum force exerted on the load is 50N

3 0
3 years ago
The wavelength of yellow sodium light in air is 589 nm.
julsineya [31]

Explanation:

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3 0
3 years ago
2. An alternating current is represented by the equation I=20sin 100mt.
Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

I=I_o sin\omega t

Where

I₀ is the peak value of current

\omega is angular frequency

As\ \omega=2\pi f

So,

f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz

We know that,

I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0
3 years ago
Speed (mph) 58 72 55 65 70 81 66 What car goes the fastest? Put the car speed in order, from fastest to slowest miles per hour.
RoseWind [281]

Answer:

The answer to your question is letter B

Explanation:

Data

Order from fastest to slowest speed

           58 72 55 65 70 81 66

The fastest is  81, then, 72, 70, then, 66, 65 finally 58 and 55

The correct order

       81 mi/h, 72 mi/h, 70 mi/h, 66 mi/h, 65 mi/h, 58 mi/h, 55 mi/h

The other options are from slowest to fastest or a different order.

7 0
3 years ago
A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
3 years ago
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