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Oliga [24]
3 years ago
5

A light-rail commuter train blows its 200 Hz horn as it approaches a crossing. The speed of sound is 339 m/s. (a) An observer wa

iting at the crossing receives a frequency of 220 Hz. What is the speed of the train? m/s (b) What frequency does the observer receive as the train moves away? Hz
Physics
1 answer:
ddd [48]3 years ago
3 0

Answer:

a) 30.82 m/s

b)  183.33 Hz

Explanation:

a)

V = speed of the sound = 339 m/s

v = speed of the train = ?

f' = observed frequency by the observer = 220 Hz

f = actual frequency of the observer = 200 Hz

using the equation

f' = \frac{fV}{V - v}

220 = \frac{(200)(339)}{339 - v}

v =  30.82 m/s

b)

V = speed of the sound = 339 m/s

v = speed of the train = 30.82 m/s

f'' = observed frequency by the observer as train moves away = ?

f = actual frequency of the observer = 200 Hz

using the equation

f'' = \frac{fV}{V + v}

f'' = \frac{(200)(339)}{339 + 30.82}

f'' = 183.33 Hz

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Explanation:

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Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
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Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

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SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

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 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

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c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

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Hope this answers the question. Have a nice day.
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