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Alenkinab [10]
2 years ago
7

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

equency of 346 Hz. (a) What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
Physics
1 answer:
aleksandr82 [10.1K]2 years ago
4 0

Answer:

Explanation:

We shall apply Doppler's effect to solve the problem .

Formula for apparent frequency for a source of sound approaching an observer is as follows .

f₁ = f₀ V / (V - v )

where f₁ and f₀ are apparent and real frequency of source , V and v is velocity of sound and velocity of approaching source respectively .

Putting the given values and knowing that speed of sound is 340 m /s

f₁ =346x 340 / (340 - 39.6 )

f₁ = 391.6 Hz

In case of receding train , the formula is

f₂ = f₀ V / (V + v )

Putting the values

f₂ = 346x 340 / (340 + 39.6 )

= 309.9 Hz

Change in frequency =  391.6 - 309.9

= 81.7 Hz .

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A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell
Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem  we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0
3 years ago
Does wavelength affect the energy of a wave?
Galina-37 [17]

Answer:

Not quite

Explanation:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength

What determines the strength of a wave?

Wave height is affected by wind speed, wind duration (or how long the wind blows), and fetch, which is the distance over water that the wind blows in a single direction. If wind speed is slow, only small waves result, regardless of wind duration or fetch.

So,

As Wavelength increases, The energy of the wave spreads and it decreases

4 0
2 years ago
Dallas says that any change in velocity is directly proportional tothe time interval over which the change took place. Dana says
Anna35 [415]

Answer:

e. Only(a) and (b) above are correct

Explanation:

Impulse

= Fx t = m ( v-u )

v-u = change in velocity

F x t = mass x change in velocity

change in velocity = F t / mass

=a t

change in velocity ∝ t ( time ) , if a is constant

dv = a_avg  dt

∫dv = a_avg ∫dt

v-u = a_avg t

change in velocity ∝ t ( time )

So both (a) and (b) are correct.

7 0
3 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
A piling for a high-rise building is pushed by two bulldozers at exactly the same time. One bulldozer exerts a force of 1250 pou
Ierofanga [76]

Answer:

Option d is correct.

The magnitude of the resultant force upon the piling = 2930 lbs.

Explanation:

Resultant of a group of forces = vector sum of the forces.

Force 1 = (-1250î) lbs

Force 2 = (2650j) lbs

Resultant = (-1250î + 2650j) lbs

The magnitude of the resultant = √[(-1250)² + (2650)²] = 2930 lbs.

Or in visualization method, the two forces presented form a right angled triangle with the resultant of the two forces.

Hence, using Pythagoras theorem,

F₁² + F₂² = R²

(1250)² + (2650)² = R²

R = √[(1250)² + (2650)²] = 2930 lbs.

7 0
3 years ago
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