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Vladimir [108]
2 years ago
12

A car approaching a stationary observer emits 450. hz from its horn. if the observer detects a frequency pf 470. hz, how fast is

the car moving? the speed of sound is 343 m/s.
Physics
1 answer:
Irina-Kira [14]2 years ago
8 0
The relationship between the frequency heard by the observer in motion and the original frequency of the sound is given by (Doppler effect)
f'= \frac{v}{v-v_s}f
where
f' is the frequency heard by the observer
v is the speed of the wave (the speed of sound)
v_s is the speed of the source relative to the observer (= the speed of the car), and it is negative when the source is approaching the observer
f is the original frequency of the sound

By re-arranging the formula, we get
v_s=v( 1- \frac{f'}{f})
and by plugging the data of the problem into the equation, we find
v_s = (343 m/s)( 1- \frac{470 Hz}{450 Hz})=-15.2 m/s
so, the car is approaching the observer at 15.2 m/s.
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A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
2 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

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for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
  • According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
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In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

       = 0.3759

r = Sin^-1 0.3759

   = 22.08

   <u>≈ 22°</u>

3 0
2 years ago
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mrs_skeptik [129]
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R  as shown in the figure attached. Then we invoke Newton's Third Law:
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If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
2 years ago
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BARSIC [14]

Answer : The process of changing a property of a wave to transmit information is called Modulation.

Explanation :

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Modulation can be done in many ways like :

(1) Frequency modulation

(2) Amplitude modulation

(3) Pulse modulation

8 0
2 years ago
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