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k0ka [10]
3 years ago
12

A train moving at a constant speed is passing a stationary observer on a platform. On one of the train cars, a flute player is c

ontinually playing the note known as concert A (f = 440 Hz). After the flute has passed, the observer hears the sound with a frequency of 415 Hz. What is the speed of the train? The speed of sound in air is 343 m/s.
Physics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

Explanation:

We shall apply the formula of Doppler effect here

F( APPARENT) = F( REAL ) X V/(V + Vs) [ v is velocity of sound and Vs is velocity of source.

415 = 440 X 343/343+Vs

142345 + 415Vs = 150920

415 V₀ = 8575

V₀ = 20.66 m/s.

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Calculate the acceleration of the car which moves 36m/s in 8s
Semenov [28]

Answer:

the answer to this question is 288

7 0
3 years ago
An apple falls straight down from a tree and hits the ground in approximately 0.75 seconds. Based on this information, which is
Mnenie [13.5K]

Answer:

y = 2.76 [m]

Explanation:

We can find the distance of the fall of the apple using the following kinematic equation, we have to emphasize that this is a typical problem of free fall, so the initial speed is zero, then we give the initial data.

t = time = 0,75[s]

g = gravity = 9.81[m/s^2]

v0 = 0

y = v_{0}*t+0.5*g*t^{2}\\ y=0.5*(9.81)*(0.75)^{2}\\y= 2.76[m]

3 0
4 years ago
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
Naddik [55]

Answer:

a

   \lambda = 1.667 nm

b

     \theta  =  0.8681^o

Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

7 0
4 years ago
Is it possible for an atom to lose a proton and have a negative charge?<br> explain
denis-greek [22]

Answer:

Sometimes atoms gain or lose electrons. The atom then loses or gains a "negative" charge. These atoms are then called ions. Positive Ion - Occurs when an atom loses an electron negative charge it has more protons than electrons.

Explanation:

:)

5 0
3 years ago
Use the equation to answer the prompt.
zavuch27 [327]

Answer:

i think it is h 8

Explanation:

6 0
3 years ago
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