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gavmur [86]
3 years ago
7

Help with the two questions above? Correct answers?

Physics
1 answer:
LenKa [72]3 years ago
7 0

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz

The wavelength is then

\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m

The third choice "0.80m; 431Hz" is correct

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The answer is A) sunlight
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What are the dimensions of a frequency in physics
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Answer:

This is the information I can provide. I hope it helps

Explanation:

Frequency is measured in units of hertz (Hz) which is equal to one occurrence of a repeating event per second. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.

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Waves travel as "packets" of several waves. in these "packets," a wave travels at __________.
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A. the same speed as the wave energy 
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The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of -5.9
Hatshy [7]
Let u =  the speed of the car at the instant when braking begins.

The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.

Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
      u = 27.2546 m/s

Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
   = 2.7696 m/s

Answer: 2.77 m/s (nearest hundredth)

4 0
3 years ago
Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down
Lorico [155]

Answer:

sin  2θ = 1    θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

            R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

          R = vo² Sin (2 15) /9.8

          R = Vo² 0.051 m

In the table are all values ​​in two ways

Angle (θ)                  distance R (x)

 0                 0                     0

15                 0.051 Vo²        0.5 Vo²/g

30                0.088 vo²        0.866   Vo²/g

45                0.102 Vo²        1   Vo²/g

60                0.088 Vo²      0.866   Vo²/g

75                0.051 vo²        0.5   Vo²/g

90                0                     0

See graphic ( R Vs θ)  in the attached ¡, it can be done with any program, for example EXCEL

6 0
3 years ago
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